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Groups satisfying (xy)^n=x^ny^n

Posted: November 11, 2018 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: exponent semigroup, Heisenberg group, n-abelian groups, torsion-free

Definition 1. Let $n$ be an integer. A group $G$ is called $n$ –abelian 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.

A simple property of subfields of R

Posted: September 23, 2018 in Elementary Algebra; Problems & Solutions, Groups and Fields

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 $F \subseteq \mathbb{Q}(x_1, \cdots , x_n).$

Maximum order of abelian subgroups in a symmetric group

Posted: April 21, 2012 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: abelian subgroups, maximum order, symmetric group

For any set $X,$ we denote by ${\rm{Sym}}(X)$ the group of bijective maps from $X$ to $X.$

Problem. Let $\alpha : \mathbb{N} \longrightarrow [1, \infty)$ be any map which satisfies the following two conditions: $\alpha(p) \geq p$ and $\alpha(p+q) \geq \alpha(p) \alpha(q)$ for all $p,q \in \mathbb{N}.$ Let $X$ be a set with $|X|=n.$ Prove that if $G$ is an abelian subgroup of ${\rm{Sym}}(X),$ then $|G| \leq \alpha(n).$

Solution. The proof is by induction on $n.$ If $n=1,$ then $|G|=1 \leq \alpha(1).$ Let $X$ be a set with $|X|= n \geq 2$ and suppose that the claim is true for any set of size $< n.$ Let $G$ be an abelian subgroup of ${\rm{Sym}}(X).$ Clearly $gx=g(x), \ g \in G, x \in X,$ defines an action of $G$ on $X.$ Let $X_1, \ldots , X_k$ be the orbits corresponding to this action and consider two cases.

Case 1. $k=1$ : Fix an element $x_1 \in X.$ Then $X=X_1=Gx_1.$ Suppose that $g_1x_1=x_1$ for some $g_1 \in G$ and let $x \in X.$ Then $x=gx_1$ for some $g \in G.$ Thus, since $G$ is abelian, we have

$x=gx_1=gg_1x_1=g_1gx_1=g_1x.$

Hence $g_1x=x$ for all $x \in X$ and thus $g_1=1.$ So the stabilizer of $x_1$ is trivial and therefore, by the orbit-stabilizer theorem, $|G|=|X|=n \leq \alpha(n).$

Case 2. $k \geq 2$ : Let $|X_i|=n_i, \ i=1,2, \ldots, k.$ Clearly $\sum_{i=1}^k n_i=n$ and, since $k \geq 2,$ we have $n_i < n$ for all $i.$ For every $g \in G$ and $1 \leq i \leq k$ let $g_i=g|_{X_i},$ the restriction of $g$ to $X_i,$ and put

$G_i=\{g_i: \ g \in G\}.$

Then $g_i \in {\rm{Sym}}(X_i)$ and $G_i$ is an abelian subgroup of ${\rm{Sym}}(X_i).$ Thus, by the induction hypothesis

$|G_i| \leq \alpha(n_i),$

for all $i.$ Now, define $\varphi : G \longrightarrow \bigoplus_{i=1}^k G_i$ by $\varphi(g)=(g_1, g_2, \ldots , g_k)$ for all $g \in G.$ It is obvious that $\varphi$ is one-to-one and so

$|G| \leq |\bigoplus_{i=1}^k G_i|=\prod_{i=1}^k |G_i| \leq \prod_{i=1}^k \alpha(n_i) \leq \alpha(\sum_{i=1}^k n_i)=\alpha(n). \ \Box$

Remark. The map $\alpha: \mathbb{N} \longrightarrow [1, \infty)$ defined by $\alpha(p)=3^{p/3},$ for all $p \in \mathbb{N},$ satisfies both conditions in the above Problem. So if $|X|=n$ and if $G$ is an abelian subgroup of ${\rm{Sym}}(X),$ then $|G| \leq 3^{n/3}.$

Degree of an irreducible factor of composition of two polynomials

Posted: November 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: composite of two polynomials, degree of irreducible factors

Problem. Let $F$ be a field and suppose that $f(x),g(x), p(x)$ are three polynomials in $F[x].$ Prove that if both $f(x)$ and $p(x)$ are irreducible and $p(x) \mid f(g(x)),$ then $\deg f(x) \mid \deg p(x).$

Solution. Let $\mathfrak{m}$ and $\mathfrak{n}$ be the ideals of $F[x]$ generated by $f(x)$ and $p(x),$ respectively. Let $E=F[x]/\mathfrak{m}$ and $L = F[x]/\mathfrak{n}.$ Since both $f(x)$ and $p(x)$ are irreducible, $E$ and $L$ are field extensions of $F.$ Now, define the map $\varphi : E \longrightarrow L$ by $\varphi(h(x) + \mathfrak{m})=h(g(x)) + \mathfrak{n},$ for all $h(x) \in F[x].$ We first show that $\varphi$ is well-defined. To see this, suppose that $h(x) \in \mathfrak{m}.$ Then $h(x)=f(x)u(x)$ for some $u(x) \in F[x]$ and hence $h(g(x)) = f(g(x))u(g(x)) \in \mathfrak{n},$ because $p(x) \mid f(g(x)).$ So $\varphi$ is well-defined. Now $\varphi$ is clearly a ring homomorphism and, since $E$ is a field and $\ker \varphi$ is an ideal of $E,$ we must have $\ker \varphi = \{0\}.$
Therefore we may assume that $F \subseteq E \subseteq L$ and hence

$\deg p(x) = [L : F] = [L: E][E:F]=(\deg f(x))[L : E]. \ \Box$

Schur’s theorem

Posted: November 2, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: commutator subgroup, commutators, index of the center of a group, Schur's theorem

Throughout $G$ is a group with the center $Z(G)$ and the commutator subgroup $G'.$ The goal is to prove that if $G/Z(G)$ is finite, then $G'$ is finite too. We will also find an upper bound for $|G'|$ in terms of $|G/Z(G)|.$

Notation. For any $a,b \in G,$ we define $[a,b]=aba^{-1}b^{-1}$ and $a^b = bab^{-1}.$

Problem 1. Let $a,b,c \in G.$

1) $[a,b]=b^ab^{-1}, \ a^a=a, \ ba=a^bb$ and $[a,b]^c=[a^c,b^c].$

2) $c[a,b]=[a^c,b^c]c.$

3) If $[G:Z(G)]=n,$ then $[a,b]^{n+1}=[a,b^2][a^b,b]^{n-1}.$

Proof. 1) The first three identities are trivial and the fourth one is true because the map $f:G \longrightarrow G$ defined by $f(g)=cgc^{-1}=g^c$ is a group homomorphism.

2) By 1) we have $[a^c,b^c]c =[a,b]^cc=c[a,b].$

3) So $g^n \in Z(G)$ for all $g \in G,$ because $[G:Z(G)]=n.$ So $[a,b]^nb^{-1}=b^{-1}[a,b]^n$ with 1) give us

$[a,b]^{n+1}=[a,b][a,b]^n =b^ab^{-1}[a,b]^n = b^a[a,b]^n b^{-1}=b^a[a,b][a,b]^{n-1}b^{-1}$

$=b^ab^ab^{-1}[a,b]^{n-1}b^{-1}= (b^2)^ab^{-2}b[a,b]^{n-1}b^{-1}=[a,b^2](b[a,b]b^{-1})^{n-1}=[a,b^2]([a,b]^b)^{n-1}$

$=[a,b^2][a^b,b^b]^{n-1}=[a,b^2][a^b,b]^{n-1}. \ \Box$

Problem 2. Let $C=\{[a,b]: \ a,b \in G\}.$ If $[G:Z(G)]=n,$ then $|C| \leq n^2.$

Proof. Define the map $\varphi : C \longrightarrow G/Z(G) \times G/Z(G)$ by $\varphi([a,b])=(aZ(G),bZ(G)).$ If we prove that $\varphi$ is one-to-one, we are done because then $|C| \leq |G/Z(G) \times G/Z(G)|=n^2.$ So suppose that $\varphi([a,b])=\varphi([c,d]).$ Then $aZ(G)=cZ(G)$ and $bZ(G)=dZ(G)$ and hence $a^{-1}c \in Z(G)$ and $b^{-1}d \in Z(G).$ Therefore

$[a,b]=aba^{-1}b^{-1}=ab(a^{-1}c)c^{-1}b^{-1} =a(a^{-1}c)bc^{-1}b^{-1}=cbc^{-1}b^{-1} =cbc^{-1}(b^{-1}d)d^{-1}$

$=cb(b^{-1}d)c^{-1}d^{-1}=cdc^{-1}d^{-1}=[c,d]. \ \Box$

Schur’s theorem. If $[G:Z(G)]=n,$ then $\displaystyle |G'| \leq n^{2n^3}.$

Solution. Let $C=\{[a,b]: \ a,b \in G\}$ and $c \in G'.$ Then $c = c_1c_2 \ldots c_m,$ where $c_i \in C.$ We will choose the integer $m$ to be as small as possible, i.e. if $c = c_1'c_2' \ldots c_k',$ with $c_i' \in C,$ then $k \geq m.$ Now, we know from Problem 2, that the number of elements of $C$ is at most $n^2.$ So if we prove that each $c_i$ can occur at most $n$ times in $c = c_1c_2 \cdots c_m,$ then $m \leq n|C| \leq n^3$ and thus there will be at most $(n^2)^m \leq n^{2n^3}$ possible values for $c$ and the problem is solved. To prove that each $c_i$ can occur at most $n$ times in $c = c_1c_2 \ldots c_m,$ suppose, to the contrary, that, say $c_j,$ occurs $r \geq n+1$ times in the product. Then by part 2) of Problem 1, we can move each $c_j$ to the right hand side of the product to get $c = c_1'c_2' \ldots c_s' c_j^r,$ where $c_i' \in C$ and $r + s = m.$ But by part 3) of Problem 1, $c_j^{n+1}$ is a product of $n$ elements of $C$ and hence, since $c_j^r=c_j^{r-(n+1)}c_j^{n+1},$ we see that $c_j^r$ is a product of $r-1$ elements of $C.$ Therefore $c$ is a product of $s+r-1=m-1$ elements of $C,$ which contradicts the minimality of $m. \ \Box$

If $G$ is finitely generated, then the converse of Schur’s theorem is also true, i.e. if $G'$ is finite, then $G/Z(G)$ is finite too. It’s not hard, try to prove it!

Partitions of a group and normal subgroups

Posted: October 9, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: normal subgroup, partition of a group

Let $G$ be a group, $N$ a normal subgroup of $G$ and $P=\{g_iN: \ i \in I \}$ the set of cosets of $N.$ Then $P$ is a partition of $G$ and $g_iN g_jN = g_ig_j N \in P$ for all $i,j \in I.$ We’d like to consider the converse of this.

Problem. Let $G$ be a group and suppose that $P$ is a set partition of $G$ which satisfies the following condition:

$(*)$ for every $Q_1,Q_2 \in P,$ there exists $Q \in P$ such that $Q_1Q_2 \subseteq Q.$

Let $N$ be the element of $P$ which contains the identity element of $G.$ Prove that $N$ is a normal subgroup of $G$ and $P$ is the set of cosets of $N$ in $G.$

Solution. Notice that an obvious result of $(*)$ is that if $Q_1 \in P$ and $g, h \in G,$ then $gQh \subseteq Q_2$ for some $Q_2 \in P.$ Now, if $a \in N,$ then $a = a \cdot 1 \in N^2$ and so $N \subseteq N^2.$ We also have $N^2 \subseteq Q$ for some $Q \in P,$ by $(*).$ Thus $N \subseteq Q$ and so $N=Q.$ Therefore $N=N ^2,$ i.e. $N$ is multiplicatively closed. Let $a \in N.$ Then $Na^{-1} \subseteq Q$ for some $Q \in P$ and since $1 \in Na^{-1},$ we get $1 \in N \cap Q.$ Thus $Q=N$ and so $Na^{-1} \subseteq N.$ Hence $a^{-1} \in N$ and so $N$ is a subgroup. To prove that $N$ is normal, let $g \in G.$ Then, by $(*),$ there exists $Q \in P$ such that $gNg^{-1} \subseteq Q.$ But then $1 \in Q \cap N,$ because $1 \in gNg^{-1} \subseteq Q,$ and so $Q=N.$ Thus $gNg^{-1} \subseteq N,$ i.e. $N$ is normal. Finally, let $g \in G$ and choose $Q, Q_1 \in P$ such that $g \in Q$ and $gN \subseteq Q_1.$ Then $g \in Q_1 \cap Q,$ because $g \in Ng \subseteq Q_1,$ and so $Q_1=Q.$ Hence $g N \subseteq Q.$ Let $Q_2 \in P$ be such that $N \subseteq g^{-1}Q \subseteq Q_2.$ Then $1 \in Q_2 \cap N,$ because $1 \in g^{-1}Q \subseteq Q_2,$ and so $Q_2=N.$ Hence $gN=Q. \ \Box$

Groups of order p^n*(p+1) are not simple

Posted: September 23, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: Burnside's normal complement theorem, centralizer, groups of order p^n*(p+1), normalizer, simple groups, Sylow theorem

As usual, for a subgroup $H$ of a group $G,$ we denote by $N(H)$ and $C(H)$ the normalizer and the centralizer of $H$ in $G.$ If $H = \langle a \rangle,$ then we write $N(a)$ and $C(a)$ for $N(H)$ and $C(H)$ respectively.

Problem. Let $p$ be a prime number and let $G$ be a group of order $p^n(p+1), \ n \geq 1.$ Prove that $G$ is not simple.

Solution. If $G$ is abelian, there is nothing to prove. So we suppose that $G$ is non-abelian and simple and we will get a contradiction. By Sylow theorem, the number of Sylow $p$ -subgroups is $p+1$ and so $P=N(P)$ for every Sylow $p$ -subgroup $P$ of $G.$ Now, we consider two cases.

Case 1 . $n=1.$ Let $P$ be a Sylow $p$ -subgroup. Then $|P|=p$ and so $P \subseteq C(P).$ Suppose that $P \neq C(P)$ and choose $a \in C(P) \setminus P.$ Then $P \subseteq C(a).$ Let $Q=aPa^{-1}.$ We have $P \cap Q = \{1\}$ because $|P|=|Q|=p$ and $a \notin P=N(P).$ Thus $PQ \subseteq C(a)$ and so $|C(a)| \geq |PQ|=p^2$ which implies that $C(a)=G$ because $|C(a)| \mid |G|=p(p+1).$ Thus $\langle a \rangle$ is in the center of $G$ and hence it’s a normal subgroup of $G,$ contradicting our assumption that $G$ is simple. Therefore $N(P)=P=C(P)$ and we are now done by the Burnside’s normal complement theorem.

Case 2. $n \geq 2.$ The idea for this case is similar to the one we used for case 2 in this problem. Let $P$ and $Q$ be two distinct Sylow $p$ -subgroups of $G$ and put $H=P\cap Q.$ Then

$\displaystyle \frac{p^{2n}}{|H|} = |PQ| \leq |G|=p^n(p+1)$

which gives $|H|=p^{n-1}$ because $|H| \mid p^n.$ As a result, $H$ is a non-trivial normal subgroup of both $P$ and $Q.$ Therefore $PQ \subseteq N(H)$ and so $|N(H)| \geq |PQ|=p^{n+1}.$
But $|N(H)| \mid |G|=p^n(p+1)$ and so $N(H)=G,$ i.e. $H$ is a normal subgroup of $G.$ This contradicts our assumption that $G$ is simple. $\Box$