Problem 1. Let G be a group and suppose that H, K are two subgroups of G. Show that if G=H \cup K, then either H=G or K=G.

Solution. If H \subseteq K or K \subseteq H, then H \cup K=G gives K=G or H=G and we are done. Otherwise, there exist h \in H \setminus K and k \in K \setminus H. But then hk \in G \setminus H \cup K, contradiction! \Box

So, as a result, if G is a finite group and H,K are two subgroups of G with H \ne G and K \ne G, then |H \cup K| \ne |G|. That raises this question: how large could |H \cup K| get? The following problem answers this question.

Problem 2. Let G be a finite group and suppose that H, K are two subgroups of G such that H \ne G and K \ne G. Show that \displaystyle |H \cup K| \le \frac{3}{4}|G|.

Solution. Recall that \displaystyle |HK|=\frac{|H||K|}{|H \cap K|} and thus \displaystyle \frac{|H||K|}{|H \cap K|} \le |G|. Hence \displaystyle |H \cap K| \ge \frac{|H| |K|}{|G|} and so

\displaystyle \begin{aligned}|H \cup K|=|H|+|K|-|H \cap K| \le |H|+|K|-\frac{|H| |K|}{|G|} =(a+b-ab)|G|, \ \ \ \ \ \ \ \ \ (*)\end{aligned}

where \displaystyle a:=\frac{|H|}{|G|} and \displaystyle b:=\frac{|K|}{|G|}.
Now, since H \ne G and K \ne G, we have [G:H] \ge 2 and [G:K] \ge 2, i.e. \displaystyle a \le \frac{1}{2} and \displaystyle b \le \frac{1}{2}. So if we let a':=1-2a and b':=1-2b, then a' \ge 0, \ b' \ge 0 and thus

\displaystyle a+b-ab=\frac{3}{4}-\frac{a'+b'+a'b'}{4} \le \frac{3}{4}.

The result now follows from (*). \ \Box

Example 1. The upper bound \displaystyle \frac{3}{4}|G| in Problem 2 cannot be improved, i.e. there exists a group G and subgroups H, K of G such that \displaystyle |H \cup K|=\frac{3}{4}|G|. An example is the Klein-four group G=\mathbb{Z}_2 \times \mathbb{Z}_2 and the subgroups H:=\{(0,0), (1,0)\} and K:=\{(0,0),(0,1)\}. Then |G|=4 and \displaystyle |H \cup K|=3=\frac{3}{4}|G|.

Example 2. We showed in Problem 1 that a group can never be equal to the union of two of its proper subgroups. But there are groups that are equal to the union of three of their proper subgroups. The smallest example, again, is the Klein-four group

\mathbb{Z}_2 \times \mathbb{Z}_2= \{(0,0), (1,0)\} \cup \{(0,0), (0,1)\} \cup \{(0,0),(1,1)\}.

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In this post, we give a nice little application of Cayley’s theorem.

Let G be a group and let g,h \in G. If g,h are conjugate in G, i.e. g=xhx^{-1} for some x \in G, then clearly g^n=1 if and only if h^n=1. So g,h have the same order. The converse however is false, i.e. if g,h \in G have the same order, that does not imply g,h are conjugate. For example, in an abelian group, two elements are conjugate if and only if they are equal but you can obviously have distinct elements of the same order in the group, e.g. in \mathbb{Z}/3\mathbb{Z}, both non-zero elements have the same order 3.

We are now going to show that although two elements of the same order of a group might not be conjugate in the group, but they are certainly conjugate in some larger group.

Problem. Let G be a group and suppose that g,h \in G have the same order. Show that there exists a group S \supseteq G such that g,h are conjugate in S.

Solution. By Cayley’s theorem, we can embed G into the symmetric group S:=\text{Sym}(G) using the injective group homomorphism f : G \to S defined by f(x)=\sigma_x \in S, where \sigma_x: G \to G is the permutation defined by \sigma_x(y)=xy for all y \in G. So we only need to show that \sigma_g, \sigma_h are conjugate in S. Well, let |g|=|h|=n. Then the cycle decomposition of \sigma_g, \sigma_h are in the form 

\sigma_g=(y_1, gy_1, \cdots , g^{n-1}y_1)(y_2, gy_2, \cdots , g^{n-1}y_2) \cdots

and 

\sigma_h=(y_1, hy_1, \cdots , h^{n-1}y_1)(y_2, hy_2, \cdots , h^{n-1}y_2) \cdots

So \sigma_g, \sigma_h have the same cycle type and hence they are conjugate in S. \ \Box

Suppose that R \ne (0) is a ring with no proper left ideals. If R has 1, then  R is a division ring. To see this, let 0 \ne x \in R. Then Rx=R and so yx=1 for some y \in R. Since y \ne 0, we have Ry=R and hence zy=1 for some z \in R. Then x=zyx=z and so yx=xy=1 proving that R is a division ring.

But what if R doesn’t have 1 ? The following problem answers this question.

Problem. Let R \ne (0) be a ring, which may or may not have 1. Show that if R has no proper left ideals, then either R is a division ring or R^2=(0) and |R|=p for some prime number p.

Solution. Let

I:=\{r \in R: \ \ Rr=(0)\}.

Then I is a left ideal of R because it’s clearly a subgroup of (R,+) and, for s \in R and r \in I, we have Rsr \subseteq Rr =(0) and so Rsr=(0), i.e. sr \in I. So either I=(0) or I=R.

Case 1: I=R. That means sr=0 for all r,s \in R or, equivalently, R^2=(0). Thus every subgroup of (R,+) is a left (in fact, two-sided) ideal of R. Hence (R,+) has no proper subgroup (because R has no proper left ideals) and thus |R|=p for some prime p.

Case 2: I=(0). Choose 0 \ne r \in R. So r \notin I and hence Rr=R because Rr is clearly a left ideal of R. Thus there exists e \in R such that er=r. Now

\text{ann}(r):=\{s \in R: \ sr=0\},

the left-annihilator of r in R, is obviously a left ideal of R and we can’t have \text{ann}(r)=R because then Rr=(0). So \text{ann}(r)=(0). Since

(re-r)r=rer-r^2=r^2-r^2=0,

we have re-r \in \text{ann}(r)=(0). Thus re=er=r. Let

J=\{x \in R: \ \ xe=x\}.

Clearly J is a left ideal of R and 0 \ne r \in J. Thus J=R. So xe=x for all x \in R. Now let 0 \ne r' be any element of R. Then, by what we just proved, r'e=r'. On the other hand, by the same argument we used for r, we find e' \in R such that r'e'=e'r'=r'. Thus r'(e-e')=0, i.e. r' \in \text{ann}(e-e').
So \text{ann}(e-e') \ne (0) and hence \text{ann}(e-e')=R, i.e. R(e-e')=(0) and thus e-e' \in I=(0).
So e=e' and hence r'e=er'=r' for all r' \in R. Thus e=1_R proving that R is a division ring. \Box

Remark. The same result given in the above problem holds if R has no proper right ideals.

Example. Let p be a prime number. The ring

\displaystyle R:= \left \{\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}: \ \ \ a \in \mathbb{Z}/p\mathbb{Z}\right\} \subset M_2(\mathbb{Z}/p\mathbb{Z})

is not a division ring and it has no proper left (or right) ideals.

Definition 1. Let n be an integer. A group G is called nabelian if (xy)^n=x^ny^n for all x,y \in G.
In other words, G is n-abelian if the map f : G \to G defined by f(x)=x^n is a group homomorphism.

Definition 2. If G is both n-abelian and m-abelian, for some integers m,n, then G is also mn-abelian because then for x,y \in G we will have

(xy)^{mn}=((xy)^m)^n=(x^my^m)^n=x^{mn}y^{mn}.

So the set

\text{E}(G)=\{n \in \mathbb{Z}: \ \ (xy)^n=x^ny^n, \ \ \forall x,y \in G\},

i.e., the set of those integers n for which G is n-abelian, is a multiplicative subset of \mathbb{Z}. Clearly 0,1 \in \text{E}(G). The set \text{E}(G) is called the exponent semigroup of G.

Remark 1. Since (xy)^n=x(yx)^{n-1}y, we have (xy)^n=x^ny^n if and only if (yx)^{n-1}=x^{n-1}y^{n-1}. So a group G is n-abelian if and only if (yx)^{n-1}=x^{n-1}y^{n-1} or, equivalently, (xy)^{1-n}=x^{1-n}y^{1-n} for all x,y \in G. So a group G is n-abelian if and only if it is (1-n)-abelian. In other words, n \in \text{E}(G) if and only if 1-n \in \text{E}(G).

Example 1. Every abelian group is obviously n-abelian for all n. It is also clear that 2-abelian groups are abelian because xyxy=(xy)^2=x^2y^2=xxyy gives yx=xy. As the next two examples show, there exists a non-abelian n-abelian group for any n > 2.

Example 2. Let n \ge 3 be an odd integer and consider the Heisenberg group G:=H(\mathbb{Z}/n\mathbb{Z}), which is a non-abelian group (why?). We show that n, n+1 \in \text{E}(G), i.e. G is both n-abelian and (n+1)-abelian. To see that, let

g= \displaystyle \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \in G.

An easy induction shows that

\displaystyle g^m= \begin{pmatrix}1 & ma & mb + \frac{m(m-1)}{2}ac \\ 0 & 1 & mc \\ 0 & 0 & 1 \end{pmatrix}

for all integers m \ge 1. Thus g^n=I, the identity matrix, and so g^{n+1}=g, for all g \in G.
Hence I=(xy)^n=x^ny^n and xy=(xy)^{n+1}=x^{n+1}y^{n+1} for all x,y \in G proving that G is both n-abelian and (n+1)-abelian.

Remark 2.  Notice that, in Example 2, if n is even, then \displaystyle \frac{n(n-1)}{2} \ne 0 (as elements of \mathbb{Z}/n\mathbb{Z}) and so g^n is not always the identity element in this case. However, there’s a way to fix this, as the next example shows. But first, notice that H(\mathbb{Z}/n\mathbb{Z}) can also be viewed as the group of  triples

G:=\{(a,b,c): \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z}\}

with multiplication defined by

(a,b,c)(a',b',c')=(a+a',b+b'+ac',c+c').

In the next example, we modify the above multiplication such that we get g^n=1 for all g \in G.

Example 3. Let n \ge 3 be any integer and consider the set G:=\{(a,b,c): \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z}\}. Define multiplication in G by

(a,b,c)*(a',b',c')=(a+a',b+b'+2ac',c+c').

It’s easy to see that (G,*) is a non-abelian group. We show that n, n+1 \in \text{E}(G). Let g=(a,b,c) \in G. A quick induction shows that

g^m=(ma, mb+m(m-1)ac, mc)

for all integers m \ge 1. So g^n=(0,0,0)=1_G for all g \in G and thus 1_G=(xy)^n=x^ny^n for all x,y \in G proving that G is n-abelian.
Also, since g^{n+1}=g for all g \in G, we have xy=(xy)^{n+1}=x^{n+1}y^{n+1} and so G is (n+1)-abelian.

Problem 1. Let G be a group. Show that

i) if n, n+1 \in \text{E}(G) for some integer n, then x^n \in Z(G) for all x \in G

ii) if n, n+1,n+2 \in \text{E}(G) for some integer n, then G is abelian.

Solution. i) Let x,y \in G. We have

xyx^ny^n=xy(xy)^n=(xy)^{n+1}=x^{n+1}y^{n+1}

and so yx^n=x^ny.

ii) Let x,y \in G. By i), we have yx^n=x^ny and yx^{n+1}=x^{n+1}y. Thus

x^{n+1}y=yx^{n+1}=yx^nx=x^nyx

and so xy=yx. \ \Box

Problem 2. Let G be an n-abelian group and let x,y \in G. Show that

i) x^{n-1}y^n=y^nx^{n-1}

ii) (xyx^{-1}y^{-1})^{n(n-1)}=1

Solution. i) By Remark 1, (ab)^{n-1}=b^{n-1}a^{n-1} for all a,b \in G. Thus

\begin{aligned} x^{n-1}y^{n-1}=(yx)^{n-1}=((yxy^{-1})y)^{n-1}=y^{n-1}(yxy^{-1})^{n-1}=y^{n-1}yx^{n-1}y^{-1}=y^nx^{n-1}y^{-1} \end{aligned}

and the result follows.

ii) Again, using the Remark 1, we have

\begin{aligned} (xyx^{-1}y^{-1})^{n(n-1)}=((x(yx^{-1}y^{-1}))^{n-1})^n=((yx^{-1}y^{-1})^{n-1}x^{n-1})^n=(yx^{-(n-1)}y^{-1}x^{n-1})^n \\ =y^n(x^{-(n-1)}y^{-1}x^{n-1})^n=y^nx^{-(n-1)}y^{-n}x^{n-1}=1, \ \ \ \text{by i)}.  \ \Box \end{aligned}

Remark 3. Let G be an n-abelian group for some integer n \ge 2. An immediate result of Problem 2, ii), is that if either G is torsion-free or G is finite and \text{gcd}(n(n-1), |G|)=1, then G is abelian.

All rings in this post are assumed to have the multiplicative identity 1.

In this post, we showed that the ring R:= \mathbb{Z}/n\mathbb{Z} has 2^m idempotents, where m is the number of prime divisors of n. Clearly m is also the number of prime (= maximal) ideals of R (recall that, in general, in commutative Artinian rings, prime ideals are maximal). Let \varphi be the Euler’s totient function. The number of units of R is \varphi(n). If n=\prod_{i=1}^m p_i^{n_i} is the prime factorization of n, then \varphi(n)=\prod_{i=}^m(p_i^{n_i}-p_i^{n_i-1}). If n is odd, then p_i^{n_i}-p_i^{n_i-1} is even for all i and hence 2^m divides \varphi(n). So if n is odd, then the number of idempotents of R divides the number of units of R. As we are going to show now, this is a property of any finite commutative ring of odd order.

Recall that a commutative ring R is called semilocal if the number of maximal ideals of R is finite and it is called local if it has only one maximal ideal.

Example 1. If p is a prime and k \ge 1 is an integer, then \mathbb{Z}/p^k\mathbb{Z} is a local ring with the unique maximal ideal p\mathbb{Z}/p^k\mathbb{Z}. If n \ge 2 is an integer, then \mathbb{Z}/n\mathbb{Z} is a semilocal ring (what are its maximal ideals?).

Example 2. Generalizing the above example, Artinian rings are semilocal. This is easy to see; let R be an Artinian ring and let S be the set of all finite intersections of maximal ideals of R. Since R is Artinian, S has a minimal element I:=\bigcap_{i=1}^m M_i. Let M be any maximal ideal of R. Since I \bigcap M \in S and I is a minimal element of S, we must have M_1M_2 \cdots M_m \subseteq I \subseteq M. So M_i \subseteq M for some i and hence M_i=M. Therefore M_1, \cdots , M_m are all the maximal ideals of R.

Problem 1. Show that if R is a local ring, then 0,1 are the only idempotents of R.

Solution. Let M be the maximal ideal of R and let e be an idempotent of R. Then e(1-e)=0 \in M. Thus either e \in M or 1-e \in M. If e \in M, then 1-e \notin M because otherwise 1=e+(1-e) \in M, which is false. So 1-e is a unit because there’s no other maximal ideal to contain 1-e. So (1-e)r=1 for some r \in R and thus e=e(1-e)r=0. Similarly, if 1-e \in M, then e \notin M and thus e is a unit. So er=1 for some r \in R implying that 1-e=(1-e)er=0 and hence e=1. \ \Box

Problem 2 Show that the number of idempotents of a commutative Arinian ring R is 2^m, where m is the number of maximal ideals of R.

Solution. Since R is Artinian, it has only finitely many maximal ideals, say M_1, \cdots , M_m (see Example 2). The Jacobson radical of R is nilpotent, hence there exists an integer k \ge 1 such that

(0)=\left(\bigcap_{i=1}^m M_i \right)^k=\prod_{i=1}^m M_i^k=\bigcap_{i=1}^m M_i^k.

Thus, by the Chinese remainder theorem for commutative rings, R \cong \prod_{i=1}^m R/M_i^k. Since each R/M_i^k is a local ring, with the unique maximal ideal M_i/M_i^k, it has only two idempotents, by Problem 1, and so R has exactly 2^m idempotents. \Box

Problem 3. Let R be a finite commutative ring. Show that if |R|, the number of elements of R, is odd, then the number of idempotents of R divides the number of units of R.

Solution. Since R is finite, it is Artinian. Let \{M_1, \cdots , M_m\} be the set of maximal ideals of R. By problem 2, the umber of idempotents of R is 2^m and R \cong \prod_{i=1}^m R/M_i^k for some integer k \ge 1.
Since |R| is odd, each |M_i| is odd too because (M_i,+) is a subgroup of (R,+) and so |M_i| divides |R|. Also, units in a local ring are exactly those elements of the ring which are not in the maximal ideal. So the number of units of each R/M_i^k is |R/M_i^k|-|M_i/M_i^k|, which is an even number because both |R| and |M_i| are odd. So the number of units of R, which is the product of the number of units of R/M_i^k, \ 1 \le i \le m, is divisible by 2^m, which is the number of idempotents of R. \ \Box

Remark 1. The result given in Problem 3 is not always true if the number of elements of the ring is even. For example, \mathbb{Z}/2\mathbb{Z} has one unit and two idempotents. However, the result is true in \mathbb{Z}/2^n\mathbb{Z}, \ n \ge 2, which has \varphi(2^n)=2^n-2^{n-1} units and two idempotents.
Can we find all even integers n for which the result given in Problem 3 is true in \mathbb{Z}/n\mathbb{Z}? Probably not because this question is equivalent to finding all integers n such that 2^m \mid \varphi(n), where m is the number of prime divisors of n, and that is not an easy thing to do.

Remark 2. The result given in Problem 3 is not necessarily true in noncommutative rings with an odd number of elements. For example, consider R:=M_2(\mathbb{F}_3), the ring of 2\times 2 matrices with entries from the field of order three. Then R has (3^2-3)(3^2-1)=48 units (see Problem 3 in this post!) but, according to my calculations, R has 14 idempotents and 14 does not divide 48.

Problem. Let R be a finite ring with 1 and let R^*=R \setminus \{0\}. Show that for every integer n \ge 1 there exist x,y,z \in R^* such that x^n+y^n=z^n if and only if R is not a division ring.

Solution. Suppose first that R is a division ring. Then, since R is a finite ring, R is a finite field, by the Wedderburn’s little theorem. Let |R|=q. Then x^{q-1}=1 for all x \in R^* and so x^{q-1}+y^{q-1}=z^{q-1} has no solution in R^*.
Conversely, suppose that R is not a division ring (equivalently, a field because R is finite). So |R| > 2 and hence the equation x+y=z has solutions in R^* (just choose y=1, \ z \ne 0,1 and x=z-1).
Let J(R) be the Jacobson radical of R. Since R is finite, it is Artinian and so J(R) is nilpotent.
So if J(R) \neq (0), then there exists a \in R^* such that a^2=0 and so a^n=0 for all n \ge 2. Therefore the equation x^n+y^n=z^n, \ n \ge 2, has a solution x=a, y=z=1 in R^*.
If J(R)=(0), then by the Artin-Wedderburn’s theorem,

\displaystyle R=\prod_{i=1}^k M_{n_i}(F_i),

for some finite fields F_i. Since R is not a field, we have either n_i >1 for some i or n_i=1 for all i and k \ge 2. If n_i > 1 for some i, then M_{n_i}(F_i), and hence R, will have a non-zero nilpotent element and we are done. If n_i=1 for all i and k \ge 2, then

x=(1,0,0, \cdots ,0), \ y = (0,1,0, \cdots , 0), \ z = (1,1,0, \cdots , 0)

will satisfy x^n+y^n=z^n. \ \Box

Problem.  Show that if \displaystyle F \subseteq \mathbb{R} is a field and x_1, \cdots , x_n are positive real numbers such that \sum_{i=1}^n x_i \in F and x_i^2 \in F for all i, then x_i \in F for all i.

Solution. The proof is by induction over n. There’s nothing to prove for n=1. For n\ge 2, since \sum_{i=1}^n x_i \in F, we have

\displaystyle \sum_{i=2}^n x_i \in F(x_1)=F + Fx_1

and thus, by the induction hypothesis, x_i \in F+ Fx_1 for all i \ge 2. So x_2=a+bx_1 for some a,b \in F.

If b=0, then x_2 \in F and so \sum_{i \ne 2}x_i \in F. Hence, by the induction hypothesis, x_i \in F for all i.

If a=0, then x_2=bx_1 (so b > 0 because x_1,x_2 > 0). Thus

\displaystyle \sum_{i=1}^n x_i = (b+1)x_1 + \sum_{i \ge 3} x_i \in F

and so, by the induction hypothesis, x_i \in F for all i.

If a, b \ne 0, then x_2^2=a^2+b^2x_1^2+2abx_1 gives x_1 \in F and so \sum_{i \ge 2} x_i \in F implying, again by the induction hypothesis,that x_i \in F for all i. \ \Box

Example. Show that \mathbb{Q}(\sqrt{2}+ \sqrt{5}+ \cdots +\sqrt{n^2+1})=\mathbb{Q}(\sqrt{2}, \sqrt{5}, \cdots , \sqrt{n^2+1}). for all integers n \ge 1.

Solution. Let F:=\mathbb{Q}(x_1 + \cdots + x_n), where x_i=\sqrt{i^2+1} for all i. Since \sum_{i=1}^n x_i \in F and x_i^2 \in F for all i, we have, by the above problem, x_i \in F for all i. Thus \mathbb{Q}(x_1, \cdots , x_n) \subseteq F and we are done because obviously F \subseteq \mathbb{Q}(x_1, \cdots , x_n).