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Finite direct product of normal subgroups with trivial intersections

Posted: August 6, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: direct product, normal subgroup

Problem. Let $H_1, H_2, \cdots , H_k, \ \ k \ge 2,$ be some normal subgroups of a group $G$ and suppose that $H_i \cap H_j=\{1\}$ for all $i \ne j.$ Show that $G$ contains a subgroup isomorphic to $H_1 \times H_2 \times \cdots \times H_k$ if $k=2,$ but not necessarily if $k \ge 3.$

Solution. Suppose first that $k=2.$ Since $H_1,H_2$ are normal, $H_1H_2$ is a subgroup of $G$ and

$h_1h_2h_1^{-1}h_2^{-1} \in H_1 \cap H_2=\{1\},$

for all $h_1 \in H_1, h_2 \in H_2$ implying that $h_1h_2=h_2h_1.$ Thus the map

$f: H_1 \times H_2 \to H_1H_2$

defined by $f(h_1,h_2)=h_!h_2$ is an onto group homomorphism. Since $H_1 \cap H_2=\{1\}, \ f$ is also one-to-one, hence an isomorphism.

A counter-example for $k \ge 3$ is $G=\mathbb{Z}_2^{k-1},$ the direct product of $k-1$ copies of $\mathbb{Z}_2.$ Clearly every $1 \ne g \in G$ has order two. Now choose $k$ distinct elements

$g_1, \cdots , g_k \in G\setminus \{1\}$

(we can do that because $k < 2^{k-1}=|G|$ ) and let $H_i$ be the subgroup of $G$ generated by $g_i.$ Since $G$ is abelian, each $H_i$ is normal in $G$ and clearly $H_i \cap H_j=\{1\}$ for all $i \ne j$ because $H_i,H_j$ are distinct subgroups of order two. Now $H_1 \times H_2 \times \cdots \times H_k$ has $2^k > 2^{k-1}=|G|$ elements and so it can’t be isomorphic to a subgroup of $G. \ \Box$

Finite subgroups of even order > 2 of GL(n,C) are not simple

Posted: May 1, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: general linear group, groups of even order, simple groups

Problem. Let $G$ be a subgroup of $\text{GL}(2, \mathbb{C}).$ Show that if $|G|$ is even and $|G| > 2,$ then $G$ is not simple.

Solution. First notice that we are done if $G$ is abelian because then every subgroup of $G$ would be normal and since $|G| > 2$ is even, $G$ has a subgroup of order two, which is clearly neither the trivial subgroup nor $G.$
So suppose, to the contrary, that $G$ is non-abelian and simple and consider the group homomorphism $f: G \to \mathbb{C}^{\times}$ defined by $f(g)=\det(g),$ for all $g \in G.$ Then since $\ker f$ is a normal subgroup of $G$ and $G$ is not simple, either $\ker f=\{I_2\},$ where $I_2$ is the identity matrix in $\text{GL}(2, \mathbb{C}),$ or $\ker f=G.$ But we can’t have $\ker f=\{I_2\}$ because then $G$ would be embedded into $\mathbb{C}^{\times}$ making $G$ abelian. Hence $\ker f=G,$ i.e. $\det g=1$ for all $g \in G.$ Now let $h \in G$ be an element of order two. So $h^2=I_2,$ which implies that $h$ is diagonalizable (because its minimal polynomial divides $x^2-1$ and so it splits into distinct linear factors). So

$h=P \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} P^{-1}$

for some $P \in \text{GL}(2, \mathbb{C}).$ Since $h^2=I_2,$ the set of eigenvalues of $h,$ which is $\{a,b\},$ is a subset of $\{-1,1\}.$ Since $\det h=1,$ we must have $a=b=\pm 1$ and so $h=\pm I_2.$ Since $h$ has order two, $h \ne I_2$ and hence $h=-I_2.$ Since $h=-I_2$ is a central element of $G,$ the subgroup $H:=\langle h \rangle$ is a normal subgroup of order two in $G$ and $H \ne G$ because $|G| > 2.$ So $G$ is not simple and that contradicts our assumption that $G$ is simple. So our assumption is wrong and $G$ is not simple indeed! $\Box$

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: general linear group, p-group, triangular

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: finite field, sum of cubes, sum of squares

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: Klein four-group, union of subgroups

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! $\Box$

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: Cayley's theorem, conjugate, symmetric group

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