Problem 1. For an integer n \ge 2, let a_n be the number of elements of order two in the symmetric group S_n. It is clear that a_1=0. Show that for any integer n \ge 2 we have

\displaystyle a_n=n!\sum_{k=1}^{\lfloor \frac{n}{2} \rfloor}\frac{1}{2^kk!(n-2k)!}.

Solution. We know that every element of S_n can be written uniquely as a product of disjoint cycles (here “uniquely” means up to the order in which the cycles are written down). We also know that the order of a product of disjoint cycles is the least common divisor of the order of the cycles. So an element of S_n has order two if and only if it’s a product of disjoint cycles of length two and that’s the number we need to find.
Let

\displaystyle \sigma=(a_1 \ a_2)(a_3 \ a_4) \cdots (a_{2k-1} \ a_{2k})

be a product of disjoint cycles of length two. Then obviously 1 \le k \le \lfloor \frac{n}{2} \rfloor. Now, there are \displaystyle \binom{n}{2} ways to choose (a_1 \ a_2) and \displaystyle \binom{n-2}{2} ways to choose \displaystyle (a_3 \ a_4) and \displaystyle \binom{n-4}{2} ways to choose \displaystyle (a_5 \ a_6) and … finally \displaystyle \binom{n-2k+2}{2} ways to choose \displaystyle (a_{2k-1} \ a_{2k}). So there are

\displaystyle \frac{1}{k!}\binom{n}{2}\binom{n-2}{2} \binom{n-4}{2} \cdots \binom{n-2k+2}{2}

possibilities for \sigma and hence

\displaystyle a_n=\sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} \frac{1}{k!}\binom{n}{2}\binom{n-2}{2} \binom{n-4}{2} \cdots \binom{n-2k+2}{2}=\sum_{k=1}^{\lfloor \frac{n}{2} \rfloor}\frac{n!(n-2)!(n-4)! \cdots (n-2k+2)!}{2^kk!(n-2)!(n-4)!(n-6)! \cdots (n-2k)!}

\displaystyle =n!\sum_{k=1}^{\lfloor \frac{n}{2} \rfloor}\frac{1}{2^kk!(n-2k)!}. \ \Box

Problem 2. For any integer n \ge 1, let b_n be the number of solutions of the equation x^2=1 in the symmetric group S_n, where 1 is the identity element of S_n. It is clear that b_1=1 and b_2=2. Show that for any integer n \ge 2 we have

b_{n+1}=b_n+nb_{n-1}.

Solution. It is clear that b_1=1, \ b_2=2. For n \ge 2, let B_n:=\{x \in S_n: \ x^2=1\}. So b_n=|B_n|. Let \sigma \in B_{n+1} and put \sigma(n+1)=k. If k=n+1, then there is b_n possibilities for \sigma. But if 1 \le k \le n, then, since \sigma^2=1, we have n+1=\sigma^2(n+1)=\sigma(k) and hence there will be b_{n-1} possibilities for \sigma.
So b_{n+1}=b_n+nb_{n-1}. \ \Box

Problem 3. Let \{a_n\},\{b_n\} be the sequences defined in Problem 1 and Problem 2, respectively. It is clear that b_n=a_n+1 and so since, by Problem 2, \{b_n\} satisfies the recurrence relation b_{n+1}=b_n+nb_{n-1}, for all n \ge 2, the sequence \{a_n\} satisfies

a_{n+1}=a_n+na_{n-1}+n \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)

for all n \ge 2. Prove (*) using the formula for a_n given in Problem 1.

Solution. I prove (*) for n even and I leave the proof for odd values of n to the reader. So let n=2m, where m is a positive integer. Then, by Problem 1,

\displaystyle \begin{aligned} a_{2m+1}=(2m+1)!\sum_{k=1}^{m}\frac{1}{2^kk!(2m-2k+1)!}=(2m)!\sum_{k=1}^{m}\frac{2m+1}{2^kk!(2m-2k+1)!}=(2m)!\sum_{k=1}^{m}\frac{2m-2k+1+2k}{2^kk!(2m-2k+1)!}\end{aligned}

\displaystyle =(2m)!\sum_{k=1}^m\frac{1}{2^kk!(2m-2k)!}+(2m)!\sum_{k=1}^m\frac{1}{2^{k-1}(k-1)!(2m-2k+1)!}

\displaystyle =a_{2m}+(2m)!\sum_{k=0}^{m-1}\frac{1}{2^kk!(2m-2k-1)!}=a_{2m}+(2m)!\left(\sum_{k=1}^{m-1}\frac{1}{2^kk!(2m-2k-1)!}+\frac{1}{(2m-1)!}\right)

=a_{2m}+2ma_{2m-1}+2m. \ \Box

Exercise. In the solution of Problem 1, to find the number of possibilities for \sigma, explain why we divided \displaystyle \binom{n}{2}\binom{n-2}{2} \binom{n-4}{2} \cdots \binom{n-2k+2}{2} by k!.

Throughout this post, F is a field and R:=F[x] is the ring of polynomials in the variable x over F.

We know that f,g \in R are coprime if and only if there exist u,v \in R such that uf+vg=1. Are u,v unique? Absolutely not, in fact there are infinitely many pairs (u,v) that satisfy uf+vg=1 because if (u,v) is one of those pairs and w \in R, then (u+wg,v-wf) is one of those pairs too. But under some conditions, we will have uniqueness, as the following problem shows.

Problem. If f,g \in R are non-constant and coprime, then there exist unique u,v \in R such that uf+vg=1 and \deg u < \deg g, \ \deg v < \deg f.

Solution. Since f,g are coprime, there exist u_1,v_1 \in R such that u_1f+v_1g=1. Dividing u_1, v_1 by g,f, respectively, gives u,v,r,s \in R that satisfy the following conditions

i) u_1=rg+u and v_1=sf+v

ii) u=0 or \deg u < \deg g and v=0 or \deg v < \deg f.

Thus

(r+s)fg + uf+vg=(u_1-u)f+(v_1-v)g+uf+vg=u_1f+v_1g=1. \ \ \ \ \ \ \ \ \ (*)

Now, by ii), either uf=0 or \deg(uf) < \deg (fg) and, similarly, either vg=0 or \deg(vg) < \deg(fg). So if r+s \ne 0, then \deg((r+s)fg + uf+vg) \ge \deg(fg) > 0 and that contradicts (*). Therefore r+s=0 and hence uf+vg=1 and that proves the existance of u,v as required. Note that since f,g are non-constant, u,v are non-zero.
To prove the uniqueness of u,v, suppose that u'f+v'g=1 for some u',v' \in R with \deg u' < \deg g and \deg v' < \deg f. Then (u-u')f=(v'-v)g and thus, since f,g are coprime, we must have f \mid v'-v, which is impossible unless v'=v because \deg(v'-v) < \deg f. So v'=v and hence (u-u')f=0, which gives u=u' and that proves the uniqueness of u,v. \ \Box

Recall that, given groups G_1,G_2, the set of group homomorphisms G_1 \to G_2 is denoted by \text{Hom}(G_1,G_2). If G_1=G_2=G, then a group homomorphism G \to G is called an endomorphism and, in this case, we write \text{End}(G) instead of \text{Hom}(G,G).

Problem. i) Let G_1,G_2 be finite abelian groups. Find |\text{Hom}(G_1,G_2)|.

ii) Let G be a finite abelian group. Find \text{End}(G) and show that |G| divides |\text{End}(G)|.

iii) Let G be a finite abelian group of order n. Show that |\text{End}(G)|= n if and only if G \cong \mathbb{Z}_n.

Solution. i) By the fundamental theorem for finite abelian groups, we have

G_1 \cong \mathbb{Z}_{m_1} \times \cdots \times \mathbb{Z}_{m_k}, \ \ \ G_2 \cong \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_{\ell}}

for some integers m_1, \cdots, m_k,n_1, \cdots, n_{\ell} \ge 2. Then by the problems in this post and this post, we have

\displaystyle \text{Hom}(G_1,G_2) \cong \text{Hom}(\mathbb{Z}_{m_1} \times \cdots \times \mathbb{Z}_{m_k},\mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_{\ell}}) \cong \prod_{1 \le i \le k, \ 1 \le j \le \ell}\text{Hom}(\mathbb{Z}_{m_i},\mathbb{Z}_{n_j})

\displaystyle \cong \prod_{1 \le i \le k, \ 1 \le j \le \ell}\mathbb{Z}_{\gcd(m_i,n_j)}.

and so, by the problem is this post,

\displaystyle |\text{Hom}(G_1,G_2)|=\prod_{1 \le i \le k, \ 1 \le j \le \ell}\gcd(m_i,n_j).

That completes the solution of i).

Now let G be a finite abelin group. Then, by the fundamental theorem for finite abelian groups,

G \cong \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k},

for some integers n_1, \cdots, n_k \ge 2, where n_i \mid n_{i+1} for all i=1, \cdots , k-1. We can now solve ii) and iii).

ii) By i), we have

\displaystyle \text{End}(G) =\text{Hom}(G,G) \cong  \prod_{1 \le i,j \le k}\mathbb{Z}_{\gcd(n_i,n_j)}

and so

\displaystyle |\text{End}(G)|=\prod_{1 \le i,j \le k}\gcd(n_i,n_j).

Thus, since n_i \mid n_j whenever i \le j, we have

|\text{End}(G)|=n_1^k(n_1n_2^{k-1})(n_1n_2n_3^{k-2}) \cdots (n_1n_2 \cdots n_k)=n_1^{2k-1}n_2^{2k-3} \cdots n_k. \ \ \ \ \ \ \ \ \ (*)

Since |G|=n_1n_2 \cdots n_k, it’s clear from (*) that |G| divides |\text{End}(G)|.

iii) Using (*) in ii), we have

n=|G|=n_1n_2 \cdots n_k=|\text{End}(G)|= n_1^{2k-1}n_2^{2k-3} \cdots n_k

if and only if k=1, i.e. n=n_1 and so G \cong \mathbb{Z}_{n_1}=\mathbb{Z}_n. \ \Box

Example 1. Let p be a prime number and suppose that G is a group of order p^2. Find \text{End}(G) and |\text{End}(G)|.

Solution. We know that a group of order p^2 is abelian. Thus, by the fundamental theorem for finite abelian groups, either G \cong \mathbb{Z}_{p^2} or G \cong \mathbb{Z}_p \times \mathbb{Z}_p.
So, by the second part of the above problem, if G \cong \mathbb{Z}_{p^2}, then \text{End}(G) \cong \mathbb{Z}_{p^2} hence

|\text{End}(G)|=p^2=|G|

and if G \cong \mathbb{Z}_p \times \mathbb{Z}_p, then \text{End}(G) \cong \mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_p hence

|\text{End}(G)|=p^4=|G|^2. \ \Box

Example 2. Find all finite abelian groups G for which |\text{End}(G)|=|G|^2.

Solution. Again, by the fundamental theorem for finite abelian groups,

G \cong \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k},

for some integers n_1, \cdots, n_k \ge 2, where n_i \mid n_{i+1} for all i=1, \cdots ,k-1. So, using the second part of the above problem, we want to have

n_1^2n_2^2 \cdots n_k^2=|G|^2= |\text{End}(G)|=n_1^{2k-1}n_2^{2k-3} \cdots n_k.

That obviously has no solution for k=1 and for k \ge 2, it is equivalent to n_k=n_1^{2k-3}n_2^{2k-5} \cdots n_{k-1}.
So the set of groups that satisfy the condition is

\{\mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k}: \ \ \ k \ge 2, \ \  n_1 \mid n_2 \mid \cdots \mid n_{k-1}, \ \ n_k=n_1^{2k-3}n_2^{2k-5} \cdots n_{k-1}\}. \ \Box

Exercise. Find all finite abelian groups G for which |\text{End}(G)|=|G|^3.

There are many infinite groups with this property that every element of the group has a finite order; for example, any direct product of infinitely many copies of a finite group.
Another example is the quotient group G:=(\mathbb{Q}/\mathbb{Z},+), which is the subject of the following problem.

Problem. Consider the additive group G:=\mathbb{Q}/\mathbb{Z}. Show that

i) G is isomorphic to the multiplicative group of complex roots of unity

ii) every finitely generated subgroup of G is cyclic

iii) G has infinitely many non-cyclic subgroups H_1, H_2, \cdots such that H_i \cap H_j=(0) for all i \ne j

iv) for every integer n \ge 1, \ G has a unique subgroup of order n

v) there is no unitary ring whose additive group is isomorphic to G.

Solution. i) Let U be the multiplicative group of complex roots of unity, i.e.

U=\{z \in \mathbb{C}: \ \ z^n=1, \ \ \text{for some integer} \ n \ge 1\}

and define the map f: G \to U by f(q+\mathbb{Z})=e^{2\pi q i}, for all q \in \mathbb{Q}. See that f is a group isomorphism.

ii) Let H be a subgroup of G generated by

\displaystyle \frac{a_i}{b_i}+\mathbb{Z}, \ \ a_i,b_i \in \mathbb{Z}, \ i=1, \cdots , n

and let b:=b_1b_2 \cdots b_n. Let K be the cyclic subgroup of G generated by \displaystyle \frac{1}{b}+\mathbb{Z}. Then for each i we have \displaystyle c_i:=\frac{b}{b_i} \in \mathbb{Z} and \displaystyle \frac{a_i}{b_i}+\mathbb{Z}=\frac{a_ic_i}{b}+\mathbb{Z} \in K. Thus H \subseteq K and hence H is cyclic (because every subgroup of a cyclic group is cyclic).

iii) For any prime number p let H_p be the subgroup of G generated by \displaystyle \frac{1}{p^n}+\mathbb{Z}, \ \ n=1,2, \cdots. Suppose that H_p is cyclic with the generator \displaystyle h=\sum_{i=1}^m \frac{a_i}{p^{n_i}}+\mathbb{Z}, \ \ a_i \in \mathbb{Z}. Let n:=\max\{n_1, \cdots , n_m\}. Then \displaystyle h=\frac{a}{p^n}+\mathbb{Z} for some a \in \mathbb{Z}. So since

\displaystyle \frac{1}{p^{n+1}}+\mathbb{Z} \in H_p=\langle h \rangle,

we must have \displaystyle \frac{1}{p^{n+1}}+\mathbb{Z}=\frac{k}{p^n}+\mathbb{Z} for some integer k. But then \displaystyle \frac{k}{p^n}-\frac{1}{p^{n+1}} \in \mathbb{Z} and so \displaystyle \frac{1}{p} \in \mathbb{Z}, which is absurd. So H_p is not cyclic. It’s easy to see that if p \ne q are primes, then H_p \cap H_q=(0).

iv) Let H be the subgroup of G generated by \displaystyle \frac{1}{n}+\mathbb{Z}. Clearly |H|=n. Now suppose that K is any subgroup of G with |K|=n. By ii), K is cyclic and so it has a generator \displaystyle \frac{a}{b}+\mathbb{Z}, where a,b \in \mathbb{Z}, \ \gcd(a,b)=1. Since |K|=n, we have \displaystyle \frac{na}{b} \in \mathbb{Z} and so b \mid n. So n=bc for some integer c and hence

\displaystyle \frac{a}{b}+\mathbb{Z}=\frac{ac}{n}+\mathbb{Z} \in H.

Thus K \subseteq H and hence K=H because |H|=|K|.

v) Suppose that R is a unitary ring and there exists an isomorphism of additive groups f: G \to (R,+). Let \displaystyle 1_R=f\left(\frac{a}{b}+\mathbb{Z}\right), \ \ a,b \in \mathbb{Z}, \ b > 0, and \displaystyle r:=f\left(\frac{1}{b+1}+\mathbb{Z}\right). Then

\displaystyle 0=f(a+\mathbb{Z})r=bf\left(\frac{a}{b}+\mathbb{Z}\right)r=b1_Rr=br=f\left(\frac{b}{b+1}+\mathbb{Z}\right)

and so \displaystyle \frac{b}{b+1} + \mathbb{Z}=0, i.e. \displaystyle \frac{b}{b+1} \in \mathbb{Z}, which is nonsense. This contradiction proves that G and (R,+) can’t be isomorphic. \Box

Exercise. This exercise is related to the second part of the above problem. Let H be a subgroup of \mathbb{Q}/\mathbb{Z} generated by \displaystyle \frac{a_i}{b_i}+\mathbb{Z}, \ \ a_i,b_i \in \mathbb{Z}, \ b_i > 0, \ i=1, \cdots , n, and let b:=\text{lcm}(b_1, \cdots , b_n). Is H generated by \displaystyle \frac{1}{b}+\mathbb{Z} \ ?

Problem. Let R be a finite commutative ring with 1. Show that the number of nilpotent elements of R divides the number of zero-divisors of R.

Solution. First two simple facts.

1) Every element of R is either a zero-divisor or a unit. That’s because if r \in R is not a zero-divisor, then Rr=R (because |Rr|=|R|) and so rx=1 for some x \in R.

2) If x \in R is nilpotent and y \in R is a unit, then x+y is a unit (see the lemma here).

Now let N(R),Z(R),U(R) be the sets of nilpotents, zero-divisors and units of R, respectively.
If x \in N(R), y \in Z(R), then z:=x+y \in Z(R) because if x \notin Z(R), then z \in U(R), by 1), and hence y=z-x \in U(R), by 2), and that’s not possible because then y \in Z(R) \cap U(R)=\emptyset. So for each y \in Z(R), we have

A_y:=\{x+y: \ \ x \in N(R)\} \subseteq Z(R)\}.

Also, if z \in Z(R), then z=0+z \in A_z. So

\displaystyle Z(R)=\bigcup_{y \in Z(R)}A_y. \ \ \ \ \ \ \ \ \ \ \ (*)

Now, if A_{y_1} \cap A_{y_2} \neq \emptyset, for some y_1,y_2 \in Z(R), then x_1+y_1=x_2+y_2 for some x_!,x_2 \in N(R) and that gives

y_2=(x_1-x_2)+y_1 \in A_{y_1}, \ \ y_1=(x_2-x_1)+y_2 \in A_{y_2}

hence A_{y_1}=A_{y_2}. The result now follows from (*) because clearly |A_y|=|N(R)| for all y \in Z(R). \ \Box

Exercise. Let’s see what the result in the problem means for R:=\mathbb{Z}/n\mathbb{Z}, the ring of integers modulo n. Let n=p_1^{k_1} \cdots p_m^{k_m} be the prime factorization of n and let \varphi be the Euler’s totient function. Let N(R), Z(R) be the sets of nilpotents and zero-divisors of R, respectively. Show that

\displaystyle |Z(R)|=n-\varphi(n), \ \ \ |N(R)|=\frac{n}{p_1 \cdots p_m}.

So, by the result given in the above problem, \displaystyle \frac{n}{p_1 \cdots p_m} \mid n-\varphi(n). Verify this result directly!

Throughout this post, G,G_1, \cdots , G_n are groups.

As always, for groups G,H, the set of group homomorphisms G \to H is denoted by \text{Hom}(G,H).

Notation. For any i=1, \cdots , n, we have a group homomorphism \mu_i: G_1 \times \cdots \times G_n \to G_i defined by

\mu_i(x_1, \cdots , x_n)=x_i.

Problem. Define the map

\varphi: \text{Hom}(G,G_1 \times \cdots \times G_n) \to \text{Hom}(G,G_1) \times \cdots \times \text{Hom}(G,G_n)

by

\varphi(f)=(\mu_1f, \cdots , \mu_nf)

for all f \in \text{Hom}(G,G_1 \times \cdots \times G_n). Show that

i) \varphi is one-to-one

ii) \varphi is onto

iii) if G_1, \cdots , G_n are abelian, then \varphi is a group isomorphism.

Solution. i) Suppose that \varphi(f)=\varphi(g) for some f,g \in \text{Hom}(G,G_1 \times \cdots \times G_n). Then, by the definition of \varphi, we have \mu_if=\mu_ig for all i=1, \cdots , n. Let x \in G and suppose that

f(x)=(y_1, \cdots , y_n), \ \ g(x)=(z_1, \cdots , z_n),

where y_i,z_i \in G_i. Then

y_i=\mu_i(f(x))=(\mu_if)(x)=(\mu_ig)(x)=\mu_i(g(x))=z_i

for all i. So f(x)=g(x) and hence f=g proving that \varphi is one-to-one.

ii) Let f_i \in \text{Hom}(G,G_i), \ \ i=1, \cdots , n, and define f: G \to G_1 \times \cdots \times G_n by

f(x)=(f_1(x), \cdots , f_n(x)),

for all x \in G. It’s clear that f is a homomorphism because each f_i: G \to G_i is a homomorphism. Also (\mu_i f)(x)=\mu_i(f(x))=f_i(x) for all x \in G and hence \mu_if=f_i. So

\varphi(f)=(\mu_1f, \cdots , \mu_nf)=(f_1, \cdots , f_n)

proving that \varphi is onto.

iii) By i), ii), we only need to show that \varphi is a group homomorphism. As I mentioned at the beginning of this post, since each G_i is abelian, \text{Hom}(G,G_i), \ i=1, \cdots , n, and \text{Hom}(G,G_1 \times \cdots \times G_n) are all abelian groups too.
Now let f,g \in \text{Hom}(G,G_1 \times \cdots \times G_n) and x \in G. Let

f(x)=(y_1, \cdots , y_n), \ \ g(x)=(z_1, \cdots , z_n).

Then

(\mu_ifg)(x)=\mu_i(fg(x))=\mu_i(f(x)g(x))=\mu_i(y_1z_1, \cdots , y_nz_n)=y_iz_i=(\mu_if)(x)(\mu_ig)(x)

=(\mu_if\mu_ig)(x)

and so \mu_ifg=\mu_if\mu_ig. Thus

\varphi(fg)=(\mu_1fg, \cdots , \mu_nfg)=(\mu_1f\mu_1g, \cdots , \mu_nf\mu_ng)=(\mu_1f, \cdots , \mu_nf)(\mu_1g, \cdots , \mu_ng)

=\varphi(f)\varphi(g)

and that means \varphi is a group homomorphism. \Box

Remark. If G,G_1, \cdots, G_n are all abelian groups, then by the above problem and the problem in the first part of this post, we have

\text{Hom}(G_1 \times \cdots \times G_n,G) \cong \text{Hom}(G_1,G) \times \cdots \times \text{Hom}(G_n,G)

and

\text{Hom}(G,G_1 \times \cdots \times G_n) \cong \text{Hom}(G,G_1) \times \cdots \times \text{Hom}(G,G_n).

Example. Find the number of group homomorphisms \mathbb{Z}_2 \times \mathbb{Z}_2 \to S_3 \times S_3.

Solution. By the above problem, parts i) and ii), there’s a bijection between \text{Hom}(\mathbb{Z}_2 \times \mathbb{Z}_2, S_3 \times S_3) and \text{Hom}(\mathbb{Z}_2 \times \mathbb{Z}_2, S_3) \times \text{Hom}(\mathbb{Z}_2 \times \mathbb{Z}_2, S_3). By Example 2 in this post, |\text{Hom}(\mathbb{Z}_2 \times \mathbb{Z}_2, S_3)|=10 and hence |\text{Hom}(\mathbb{Z}_2 \times \mathbb{Z}_2, S_3 \times S_3)|=|\text{Hom}(\mathbb{Z}_2 \times \mathbb{Z}_2, S_3)|^2=100. \ \Box

Throughout this post, G,G_1, \cdots , G_n are groups.

As always, for groups G,H, the set of group homomorphisms G \to H is denoted by \text{Hom}(G,H).

Notation. For any i=1, \cdots , n, we have a group homomorphism \lambda_i: G_i \to G_1 \times \cdots \times G_n defined by

\lambda_i(x_i)=(1, \cdots 1,x_i,1 \cdots , 1).

See that (x_1, \cdots , x_n)= \lambda_1(x_1) \cdots \lambda_n(x_n) for all (x_1, \cdots , x_n) \in G_1 \times \cdots \times G_n.

Problem. Define the map

\varphi: \text{Hom}(G_1 \times \cdots \times G_n,G) \to \text{Hom}(G_1,G) \times \cdots \times \text{Hom}(G_n,G)

by

\varphi(f)=(f\lambda_1, \cdots , f\lambda_n)

for all f \in \text{Hom}(G_1 \times \cdots \times G_n,G). Show that

i) \varphi is one-to-one

ii) if G is abelian, then \varphi is a group isomorphism.

Solution. i) Suppose that \varphi(f)=\varphi(g) for some f,g \in \text{Hom}(G_1 \times \cdots \times G_n,G). Then, by the definition of \varphi, we have f \lambda_i=g\lambda_i for all i=1, \cdots , n and thus for (x_1, \cdots , x_n) \in G_1 \times \cdots \times G_n, we have

f(x_1, \cdots , x_n)=f(\lambda_1(x_1) \cdots \lambda_n(x_n))=f(\lambda_1(x_1)) \cdots f(\lambda_n(x_n))=g(\lambda_1(x_1)) \cdots g(\lambda_n(x_n))

=g(\lambda_1(x_1) \cdots \lambda_n(x_n))=g(x_1, \cdots , x_n)

and so f=g proving that \varphi is one-to-one.

ii) We have already shown in i) that \varphi is injective. So we only need to show that \varphi is a an onto group homomorphism. Note that, as I mentioned at the beginning of this post, since G is an abelian group, \text{Hom}(G_i,G), \ i=1, \cdots , n, and \text{Hom}(G_1 \times \cdots \times G_n,G) are all abelian groups too.

1) \varphi is a group homomorphism. To prove this, let f,g \in \text{Hom}(G_1 \times \cdots \times G_n,G) and x_i \in G_i. Then

(f\lambda_i g \lambda_i)(x_i)=(f\lambda_i)(x_i)(g\lambda_i)(x_i)=f(\lambda_i(x_i))g(\lambda_i(x_i))=fg(\lambda_i(x_i))=(fg\lambda_i)(x_i)

and so fg\lambda_i=f\lambda_ig\lambda_i for all i=1, \cdots , n. Thus

\varphi(fg)=(fg\lambda_1, \cdots , fg\lambda_n)=(f\lambda_1g\lambda_1, \cdots , f\lambda_ng\lambda_n)=(f\lambda_1, \cdots , f\lambda_n)(g\lambda_1, \cdots, g\lambda_n)

=\varphi(f)\varphi(g).

2) \varphi is onto. To prove this, let f_i \in \text{Hom}(G_i,G), \ i=1, \cdots , n and define f: G_1 \times \cdots \times G_n \to G by

f(x_1, \cdots , x_n)=f_1(x_1) \cdots f_n(x_n)

for all x_i \in G_i, \ i=1, \cdots , n. Since each f_i: G_i \to G is a group homomorphism and G is abelian, f is a group homomorphism and

f\lambda_i(x_i)=f(1, \cdots , 1, x_i,1, \cdots, 1)=f_1(1) \cdots f_i(x_i) \cdots f_n(1)=f_i(x_i).

So f \lambda_i=f_i for all i and hence

\varphi(f)=(f\lambda_1, \cdots , f\lambda_n)=(f_1, \cdots , f_n)

proving that \varphi is onto. \Box

Example 1. i) Let G be a finite abelian group and let n \ge 2 be an integer. Find the number of elements of \text{Hom}(G,\mathbb{Z}_n) and \text{Hom}(G,\mathbb{Z}).

ii) Find the number of group homomorphisms \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_4 \to \mathbb{Z}_{10}.

Solution. i) By the fundamental theorem for finite abelian groups, we have

G=\mathbb{Z}_{m_1} \times \cdots \times \mathbb{Z}_{m_k}

for some integers m_1, \cdots , m_k \ge 2. Thus, by the above problem and the problem in this post, we have

|\text{Hom}(G,\mathbb{Z}_n)|=|\text{Hom}(\mathbb{Z}_{m_1},\mathbb{Z}_n) \times \cdots \times \text{Hom}(\mathbb{Z}_{m_k},\mathbb{Z}_n)|=\prod_{i=1}^k |\text{Hom}(\mathbb{Z}_{m_i},\mathbb{Z}_n)|

=\prod_{i=1}^k \gcd(m_i,n).

and

|\text{Hom}(G,\mathbb{Z})|=|\text{Hom}(\mathbb{Z}_{m_1},\mathbb{Z}) \times \cdots \times \text{Hom}(\mathbb{Z}_{m_k},\mathbb{Z})|=\prod_{i=1}^k |\text{Hom}(\mathbb{Z}_{m_i},\mathbb{Z})|

=\prod_{i=1}^k |(0)|=1.

ii) By i), the number is \gcd(2,10)\gcd(3,10)\gcd(4,10)=4. \ \Box

Remark. By the first part of the above problem, the map \varphi is always one-to-one whether G is abelian or not. But if G is not abelian, then \varphi need not be onto, as the next example shows.

Example 2. Let A:=\text{Hom}(\mathbb{Z}_2,S_3) and B:=\text{Hom}(\mathbb{Z}_2 \times \mathbb{Z}_2, S_3). Show that |A|=4 and |B|=10.

Solution.  A group homomorphism f: \mathbb{Z}_2 \to S_3 is determined by the value of f(1). Since 1 \in \mathbb{Z}_2 has order two, we either have f(1)=1 or f(1) is an element of order two in S_3. So since S_3 has exactly three elements of order two (what are they?), |A|=4.
To find |B|, let f: \mathbb{Z}_2 \times \mathbb{Z}_2 \to S_3 be a non-trivial group homomorphism and K:=\ker f. So |K| < 4 and since S_3 has no subgroup of order four, f can’t be injective, i.e. |K| > 1. So |K|=2. Now, write

\mathbb{Z}_2 \times \mathbb{Z}_2=\{1,x,y,xy\}

where both x,y have order two and xy=yx. There are three possibilities for K

K=\{1,x\}, \ \ K=\{1,y\}, \ \ K=\{1,xy\}.

Suppose that K=\{1,x\}. Then f(xy)=f(y) and f(y) is an element of order two in S_3. Since S_3 has exactly three elements of order two, there are exactly three ways to define f in this case. Similarly, there are three possibilities for f if K=\{1,y\} or K=\{1,xy\}. Thus there are exactly nine non-trivial group homomorphisms from \mathbb{Z}_2 \times \mathbb{Z}_2 to S_3, which means |B|=10. \ \Box

Note that, in Example 2, we have |A \times A|=16> 10=|B| and so the map \varphi, as defined in the above problem, is not onto in this case.

In the second part of this post, we study the relationship between \text{Hom}(G, G_1 \times \cdots \times G_n) and \text{Hom}(G,G_1) \times \cdots \times \text{Hom}(G,G_n).