Showing $|G|=90$ means $G$ is not simple by centraliser argument

Question

I want to show that a group $G$, with $|G|=90$ cannot be simple, specifically using a centraliser argument. The exercise gives a walkthrough really of what I am to do, but I am even then, still having trouble:

Assume $G$ is simple. Let $P$ be a Sylow $3$-subggrup of $G$ and let $x\in P, x\ne 1$. Show that $C_G(x)=P$ and obtain a contradiction to conclude that $G$ is not simple.

Now $n_3=1$ or $10$, and $P$ is of order $9=3^2$,(we know that $|H|=p$ or $p^2$, means that $H$ is abelian) which means that $P$ is abelian, hence $C_P(x)\supseteq P$, so this is where I am stuck.

I understand that if $C_G(x)=P$ then I am done, since $C_G(x)\trianglelefteq G$, but if $P\subset G$ is abelian, is then $C_G(x)=P$? i.e. is it abelian with elements of $G\backslash P$ for some reason?

Answer 1

Every group $G$ is isomorphic to a subgroup $H$ of $A(G)$ i.e the set of permutations of $G$.So $\exists \phi:G\to A(G)$ a monomorphism defined by $\phi(g)=\tau_g;\tau_g:G\to G;\tau_g=ag\forall a\in G$

$H$ has an even permutation namely identity. Also $G$ has an element of order $2$ and so does $H$ say $a$.Also $\tau(g)=ag;\tau_g(ga)=a^2g=g$

So $\tau_g$ is the product of transpositions of the form $(a,ga)$.Since $|G|=2m$ so number of transpositions in the factorisation of $\tau$ is odd and hence $\tau$ is an odd permutation.

Define $f:H\to \{-1,1\}$ by $f(x)=1 $if $x $ is even and $f(x)=-1 $ if $x$ is odd.

Then $H/ker f \equiv Im f\implies |H/kerf |=2 \implies |kerf |=m$.

Thus $H$ and hence $G$ contains a normal subgroup of order $m$

Answer 2

Assume $G$ is simple. Then the number of Sylow-3 subgroups must be $10$, let they be $$P_1, P_2, \cdots, P_{10}.$$ These subgroups are abelian groups of order $3^2$.

Case (A) $P_i\cap P_j\neq 1$ for some $i\neq j$ in $\{1,2,\cdots,10\}.$

What is $|P_i\cap P_j|$? It is not $1$ by assumption, and it divides $|P_i|=9$. So if $|P_i\cap P_j|=9$ then $P_i=P_j$, contradiction. It must be that $|P_i\cap P_j|=3$.

Consider non-identity $x$ in $P_i\cap P_j$. Then $C_G(x)$ contains $P_i$ as well as $P_j$ (hence $P_iP_j$).

But $$|P_iP_j|=\frac{|P_i|.|P_j|}{|P_i\cap P_j|}=\frac{9.9}{3}=27.$$ Thus, $|C_G(x)|\geq 27$, $|C_G(x)|$ is divisible by $|P_i|=9$ and $|C_G(x)|$ divides $|G|$, hence $|C_G(x)|$ is either $45$ or $90$.

A subgroup of order $45$ will have index $2$, so normal, contradiction.

If $C_G(x)=G$ then $Z(G)\neq 1$, contradiction (why?)

Case (B) $P_i\cap P_j=1$ for all $i\neq j$ in $\{1,2,\cdots,10\}.$

Then non-identity elements of $P_1,P_2,\cdots,P_{10}$ cover $(9-1).10=80$ elements of order $3$ or $3^2$.

Come to Sylow-$5$ subgroups. How many there can be? If it is more than $1$, then there will be at least $1+1.5=6$. And they will cover at least $(5-1).6=24$ new elements. This is contradiction (why?)?