## Finite rings in which sum and product of all non-zero elements are non-zero

Posted: March 29, 2019 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags:

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

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

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 $R=R_0 \cong \mathbb{Z}_4. \ \Box$

## Solving the system of equations yx^(n+1)=x^ny and xy^(n+1)=y^nx in a group

Posted: March 24, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields

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 $y^{n+1}=x^{-1}y^nx=y^n. \ \Box$

## A finite p-subgroup of GL(n,k)

Posted: February 27, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields, Linear Algebra
Tags: , ,

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 $p. \ \Box$

## Elements of a finite field as a sum of two squares or two cubes

Posted: February 6, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

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 1$q=2n$ 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 2$q=2n+1$ 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 $3.$

## Maximum number of elements of the union of two proper subgroups of a finite group

Posted: December 2, 2018 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

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

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

## Two elements of the same order of a group are conjugate, in somewhere!

Posted: November 24, 2018 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

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 $S. \ \Box$

## Rings with no proper left ideals

Posted: November 16, 2018 in Elementary Algebra; Problems & Solutions, Rings and Modules

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.