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 ii).

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.

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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).

Given groups G_1,G_2, the set of all group homomorphisms G_1 \to G_2 is  denoted by \text{Hom}(G_1,G_2).
Recall that if G_2 is abelian, then \text{Hom}(G_1,G_2) has a natural abelian group structure defined by

(fg)(x)=f(x)g(x), \ \ \ \ \ \forall f,g \in \text{Hom}(G_1,G_2), \ x \in G_1.

In this post, we find \text{Hom}(G_1,G_2) for cyclic groups G_1,G_2.

The important point is that if (G_1,+) is a cyclic group generated by g and if G is any group, then a group homomorphism f: G_1 \to G is completely determined by f(g) because every element of G_1 is in the form ng for some integer n and f(ng)=(f(g))^n.

Remark 1. To keep it simple, in the solution of the following problem, we write k for both k \in \mathbb{Z} and the coset k+\ell \mathbb{Z} \in \mathbb{Z}_{\ell}.

Problem. Let m, n \ge 2 be integers and let d:=\gcd(m,n). Show that

i) \text{Hom}(\mathbb{Z}_m,\mathbb{Z}_n) \cong \mathbb{Z}_d

ii) \text{Hom}(\mathbb{Z}_n,\mathbb{Z}) =(0)

iii) \text{Hom}(\mathbb{Z},\mathbb{Z}_n) \cong \mathbb{Z}_n

iv) \text{Hom}(\mathbb{Z},\mathbb{Z}) \cong \mathbb{Z}.

Solution. i) Let f \in \text{Hom}(\mathbb{Z}_m,\mathbb{Z}_n) and f(1)=k. Then 0=f(0)=f(m)=mf(1)=mk and thus n \mid mk, which gives \displaystyle \frac{n}{d} \mid k. Hence

\displaystyle k=\frac{jn}{d}, \ \ 0 \le j \le d-1.

So, since f is completely determined by k=f(1) and k has d possible values, there are d possible ways to define f. Finally, if f \in \text{Hom}(\mathbb{Z}_m,\mathbb{Z}_n) is defined by \displaystyle f(1)=\frac{jn}{d} for some 0 \le j \le d-1, then f=jg, where g \in \text{Hom}(\mathbb{Z}_m,\mathbb{Z}_n) is defined by \displaystyle g(1)=\frac{n}{d}. That means \text{Hom}(\mathbb{Z}_m,\mathbb{Z}_n)=\langle g \rangle \cong \mathbb{Z}_d.

ii) Let f \in \text{Hom}(\mathbb{Z}_n,\mathbb{Z}) and f(1)=k. Then 0=f(n)=nf(1)=nk and so k=0 implying that f=0.

iii) Let f \in \text{Hom}(\mathbb{Z},\mathbb{Z}_n). Since 1 \in \mathbb{Z} has infinite order, f(1) can be defined to be any element of \mathbb{Z}_n and if f(1)=k, then f=kg, where g \in \text{Hom}(\mathbb{Z},\mathbb{Z}_n) is defined by g(1)=1. So \text{Hom}(\mathbb{Z},\mathbb{Z}_n) =\langle g \rangle \cong \mathbb{Z}_n.

iv) Same as iii). \Box

Remark 2. If \gcd(m,n)=1, then, by the first part of the above problem, \text{Hom}(\mathbb{Z}_m,\mathbb{Z}_n) \cong \mathbb{Z}_1=(0).

As always, M_n(\mathbb{C}) is the ring of n\times n matrices with complex entries.

Problem. Let

Z:=\{A \in M_n(\mathbb{C}): \ \ \text{tr}(A)=0\}, \ \ \ \ \ N:=\{A \in M_n(\mathbb{C}): \ \ A \ \text{is nilpotent}\}.

i) Show that Z is a vector space but N is not.

ii) Show that Z=\text{span}(N).

SolutionLet \{e_{ij}: \ 1 \le i,j \le n\} \subset M_n(\mathbb{C}) be the standard basis of M_n(\mathbb{C}).

i) Let A_1,A_2 \in Z and a_1,a_2 \in \mathbb{C}. Then

\text{tr}(a_1A_1+a_2A_2)=a_1\text{tr}(A_1)+a_2\text{tr}(A_2)=0.

So a_1A_1+a_2A_2 \in Z and thus Z is a vector space. To show that N is not a vector space, see that, for example, e_{12}, e_{21} \in N, because e_{12}^2=e_{21}^2=0, but e_{12}+e_{21} \notin N because (e_{12}+e_{21})^3=e_{12}+e_{21}.

ii) Since all the eigenvalues of a nilpotent matrix are zero, the trace of a nilpotent matrix zero and thus N \subseteq Z implying that \text{span}(N) \subseteq Z, because Z is a vector space.
So, to complete the proof of ii), we need to show that Z \subseteq \text{span}(N). Let A \in Z. Let U be an upper triangular element of M_n(\mathbb{C}) similar to A (such U exists because the field of complex numbers is algebraically closed). So

A=PUP^{-1} 

for some invertible element P \in M_n(\mathbb{C}). Clearly we can write U=D+V, where D \in M_n(\mathbb{C}) is diagonal and V \in M_n(\mathbb{C}) is strictly upper triangular. Notice that V is nilpotent because all of its eigenvalues are zero (because its eigenvalues are the entries on the main diagonal and those entries are all zero). So

A=PDP^{-1}+PVP^{-1} 

and PVP^{-1} \in N. Thus, in order to complete the solution, we only need to show that D \in \text{span}(N). Clearly D \in Z because A,V \in Z. Let d_i be the (i,i)-entry of D. Since D \in Z, we have d_n=-\sum_{i=1}^{n-1}d_i and thus

\displaystyle D=\sum_{i=1}^{n-1}d_i(e_{ii}-e_{nn}).

So, in order to show that D \in \text{span}(N), we only need to show that e_{ii}-e_{nn} \in \text{span}(N) for all 1 \le i < n and to do that, just write 

\displaystyle e_{ii}-e_{nn}=\frac{1}{2}(e_{ii}+e_{in}-e_{ni}-e_{nn})+\frac{1}{2}(e_{ii}-e_{in}+e_{ni}-e_{nn}). \ \ \ \ \ \ \ \ \ \ \ (*) 

and see that 

(e_{ii}+e_{in}-e_{ni}-e_{nn})^2=(e_{ii}-e_{in}+e_{ni}-e_{nn})^2=0. 

So both e_{ii}+e_{in}-e_{ni}-e_{nn} and e_{ii}-e_{in}+e_{ni}-e_{nn} are in N and therefore e_{ii}-e_{nn} \in \text{span}(N), by (*). \ \Box