Problem. Let R be a finite ring with 1 and suppose that R satisfies the following properties

i) \displaystyle \sum_{r \in R} r \ne 0

ii) \displaystyle \prod_{0 \ne r \in R} r \ne 0.

Show that either R \cong \mathbb{Z}_2 or R \cong \mathbb{Z}_4.

Solution. First notice that both \mathbb{Z}_2 and \mathbb{Z}_2 satisfy i), ii). Since \mathbb{Z}_2 is the only ring with two elements, we may assume that |R| > 2. Let n:=|R|. First we show that n is even. Suppose, to the contrary, that n is odd. Then the group (R,+) would have no element of order 2, i.e. r \ne -r for all 0 \ne r \in R, and thus

\displaystyle \sum_{r \in R} r = \sum (r+(-r))=0,

contradicting the property i).
Now let m:=\text{char}(A). We show that m=p^k for some prime p and k \in \{1,2\}. To see that, let m=\prod_{i=1}^{\ell}p_i^{k_i} be the prime factorization of m. If \ell > 1, then

\displaystyle \prod_{i=1}^{\ell} p_i^{k_i}1_R=m1_R=0,

contradicting the property ii), because p_i^{k_i}1_R are non-zero distinct elements of R. So \ell=1, i.e. m=p^k for some prime number p and positive integer k. Now if k > 2, then we can write

0=m1_A=(p1_R)(p^{k-1}1_R)

and again, by the minimal property of m, the elements p1_R and p^{k-1}1_R are non-zero and distinct and that contradicts the property ii). So k \le 2.
We now show that in fact m=4. We just proved that m=p^k for some prime p and k \in \{1,2\}. Suppose that q \ne p is a prime divisor of n. Then there exists r \in (R,+) that has order q. But then qa=ma=0 which gives the false result r=0, because q,m are coprime. So n is a power of p. But we have already showed that n is even. So p=2 and thus either m=2 or m=4. If m=2, then R would be a vector space over \mathbb{Z}_2, i.e. R=\sum_{i=1}^s \mathbb{Z}_2r_i for some integer s \ge 2, because n > 2, and r_i \in R. But then

\displaystyle \sum_{r \in R}r =2^{s-1}\sum_{i=1}^s r_i=0,

contradicting the property ii). So m=4 and hence R contains a subring R_0 \cong \mathbb{Z}_4.
Finally, we show that R=R_0. Suppose, to the contrary, that R \neq R_0 and let r \in R \setminus R_0. We have (21_R)(2r)=0, because \text{char}(R)=4, and so, since 21_R \ne 0, we must have either 21_R=2r or 2r=0, by property ii). If 21_R=2r, then (21_R)(r-1_R)=0, which gives r=1_R \in R_0, by property ii), and that’s a contradiction. So 2r=0, i.e. we have shown that 2r=0 for any element r \in R \setminus R_0. But if r \in R \setminus R_0, then r+1_R \in R \setminus R_0 too, because 1_R \in R_0, and so 2r=2(r+1_R)=0, which gives 21_R=0, contradiction. So R=R_0 \cong \mathbb{Z}_4. \ \Box

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Problem. Let G be a group and let n be a positive integer. Find all x,y \in G that satisfy the system of equations yx^{n+1}=x^ny and xy^{n+1}=y^nx.

Solution. I show that only x=y=1 satisfy the system. First see that from x^{n+1}=y^{-1}x^ny we get

x^{(n+1)^{n+1}}=y^{-(n+1)}x^{n^{n+1}}y^{n+1} \ \ \ \ \ \ \ \ \ \ (1)

and so, since y^{n+1}=x^{-1}y^nx, we have x^{(n+1)^{n+1}}=x^{-1}y^{-n}x^{n^{n+1}}y^nx, which gives

x^{(n+1)^{n+1}}=y^{-n}x^{n^{n+1}}y^n. \ \ \ \ \ \ \ \ \ \ (2)

Now, (1) and (2) together give

yx^{n^{n+1}}=x^{n^{n+1}}y \ \ \ \ \ \ \ \ \ \ \ (3)

and so, by (1) or (2),

x^{(n+1)^{n+1}}=x^{n^{n+1}}. \ \ \ \ \ \ \ \ \ (4)

We also have from x^{n+1}=y^{-1}x^ny that x^{(n+1)^{n+1}}=y^{-1}x^{n(n+1)^n}y and so, by (3), (4),

x^{n(n+1)^n}=x^{(n+1)^{n+1}}. \ \ \ \ \ \ \ \ \ (5)

It follows from (5) that x^{(n+1)^n}=1 and thus 1=x^{(n+1)^{n+1}}=x^{n^{n+1}}, by (4). Therefore x=1, because n^{n+1} and (n+1)^{n+1} are relatively prime, and so y=1 because y^{n+1}=x^{-1}y^nx=y^n. \ \Box

Problem. Let k be a field, n a positive integer. Let G be a finite subgroup of \text{GL}(n,k) such that |G|>1 and suppose also that every g \in G is upper triangular and all the diagonal entries of g are 1.
Show that \text{char}(k) > 0 and |G| is a power of \text{char}(k).

Solution. First, let’s put an order on the set

S:=\{(i,j): \ \ 1 \le i < j \le n\}.

We write (i,j) < (i',j') if i < i' or i=i', \ j < j'. Now let g=[g_{ij}] be any non-identity element of G. Let (r,s) be the smallest element of S such that g_{rs} \ne 0. See that, for any integer m, the (r,s)-entry of g^m is mg_{rs} and the (i,j)-entries of g^m, where (i,j) \in S, \ (i,j) < (r,s), are all zero. But since G is finite, there exists an integer m > 1 such that g^m is the identity matrix and so we must have mg_{rs}=0. Thus m1_k=0 (because g_{rs} \ne 0) and hence p:=\text{char}(k) > 0. So the (i,j)-entries of g^p, where (i,j) \in S and (i,j) \le (r,s), are all zero.
Now if g^p is not the identity matrix, we can replace g with g^p and repeat our argument to find an element (u,v) \in S, \ (u,v) > (r,s), such that all (i,j)-entries of g^{p^2}, where (i,j) \in S, \ (i,j) \le (u,v), are zero. Then, again, if g^{p^2} is not the identity matrix, we repeat the argument for g^{p^2}, etc. Since g has only finitely many entries, there exists some positive integer \ell such that all the (i,j)-entries of g^{p^{\ell}}, where (i,j) \in S, are zero. That means g^{p^{\ell}} is the identity matrix and hence |g| is a power of p. So we have shown that the order of every non-identity element of G is a power of p. Thus |G| is a power of p. \ \Box

It is well-known that, in a finite field, every element is a sum of two squares (Problem 1). It is however not true that every element of a finite field is a sum of two cubes. For example, in \mathbb{Z}_7, we cannot write 3 or 4 as a sum of two cubes because \{a^3: \ a \in \mathbb{Z}_7\}=\{0,1,6\} and so the only elements of \mathbb{Z}_7 that are a sum of two cubes are 0,1,2,5,6.
But if, in a finite field, \alpha^3=2 for some non-zero element \alpha of the field, then we can show that every element of the field is a sum of two cubes (Problem 2).

Problem 1. Show that every element of a finite field is a sum of two squares.

Solution. Let F be a finite field. So we want to show that if x \in F, then x=a^2+b^2 for some a,b \in F. We can actually be more specific if we consider two cases. Let |F|=q.

Case 1q=2n for some integer n. Then, since x^q=x for all x \in F, we get x=(x^n)^2. So in this case, every element of the field is a square.

Case 2q=2n+1 for some integer n. Since F is finite, the multiplicative group F^{\times} is cyclic.
So F^{\times}=\langle c \rangle. Let x \in F and consider the sets

A:=\{a^2: \ \ a \in F\}, \ \ \ B:=\{x-a^2: \ \ a \in F\}.

Clearly \{0\} \cup \{c^{2m}: \ \ 1 \le m \le n\} \subseteq A and |A|=|B|. Thus |A| \ge n+1 and hence

|A|+|B| =2|A| \ge 2n+2 > |F|.

Therefore A \cap B \neq \emptyset, i.e. there exist a,b \in F such that a^2=x-b^2 and the result follows. \Box

Remark 1. Regarding the second case in the solution of Problem 1, notice that, in fact, we have

A= \{0\} \cup \{c^{2m}: \ \ 1 \le m \le n\}

and so |A|=n+1. The reason is that if c^k=c^{2m} for some integers k,m, then c^{k-2m}=1 and hence k-2m must be divisible by |F^{\times}|=2n implying that k is even.

Problem 2. Let F be a finite field and suppose that there exists 0 \ne \alpha \in F such that \alpha^3=2. Show that every element of F is a sum of two cubes.

Solution. So we want to show that if x \in F, then x=a^3+b^3 for some a,b \in F. Let |F|=q and let’s consider three cases.

Case 1: q=3n for some integer n. Then x=x^q=(x^{n})^3 for all x \in F.

Case 2: q=3n+2 for some integer n. Then x=x^{2-q}=(x^{-n})^3 for all 0 \ne a \in F and clearly 0=0^3.

So, in both cases 1 and 2, for every x \in F, there exists a \in F such that x=a^3.

Case 3: q=3n+1 for some integer n. Since F is finite, the multiplicative group F^{\times} is cyclic. So F^{\times}=\langle c \rangle. Let x \in F and consider the sets

A:=\{a^3: \ \ a \in F\}, \ \  B:=\{x-a^3: \ \ a \in F\}, \ \ \ C:=\{a^3-x: \ \ a \in F\}.

Clearly \{0\} \cup \{c^{3m}: \ \ 1 \le m \le n\} \subseteq A and |A|=|B|=|C|. So |A| \ge n+1 and

|A|+ |B|+ |C| =3|A| \ge 3n+3 > |F|.

So at least two of the sets A,B,C have non-empty intersection. If A \cap B \neq \emptyset or A \cap C \neq \emptyset, then x=a^3+b^3 for some a,b \in F and we are done.
Now suppose that B \cap C \ne \emptyset. So there exist a,b \in F such that x-a^3=b^3-x and so 2x=a^3+b^3. Since, as given in the problem, \alpha^3=2 for some \alpha \ne 0, we have 2 \ne 0 and 2^{-1}=\alpha^{-3}. Hence

x=\alpha^{-3}(a^3+b^3)=(\alpha^{-1}a)^3+(\alpha^{-1}b)^3. \ \Box

Remark 2. Regarding the third case in the solution of Problem 2, notice that, in fact, we have

A= \{0\} \cup \{c^{3m}: \ \ 1 \le m \le n\}

and so |A|=n+1. The reason is that if c^k=c^{3m} for some integers k,m, then c^{k-3m}=1 and hence k-3m must be divisible by |F^{\times}|=3n implying that k is divisible by 3.

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)\}.

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.