How do I show that every group of order 90 is not simple? Can comeone help me? How do I show that every group of order 90 is not simple?
 A: This is a very very late answer. But anyway I'll post for the use of people who are learning form this site (like myself).
First note that $90=2\times 3^2\times 5$ and of the form $pq^2r$ with three different primes (like $60=2^2\times 3\times 5$). Therefore this case is bit hard comparing with common exam/ homework problems, where we have the orders of the form $p, p^2, p^3, pq, pq^2, pqr.$
Assume a group $G$ of order $90$ is simple. Using Sylow theorems, with standard notations, it is not difficult to prove that $n_3=10$ and $n_5=6.$ So we have $6\times(5-1)=24$ elements of order $5$ and if $3$-Sylow subgroups intersects trivially then we have $10\times(9-1)=80$ elements of order $3,$ but this contradicts to the order of the group as $24+80\gt 90.$ Therefore $3$-Sylow subgroups must have a non-trivial intersection.
Let $H, K\in Syl_3(G),$ then $|H\cap K|=3$ and $$|HK|=\dfrac{|H||K|}{|H\cap K|}=27.$$ Further $[H :|H\cap K|]=[K:|H\cap K|]=3$ is the smallest prime factor of both $|H|, |K|.$ Therefore $H\cap K$ is normal in both $H$ and $K.$ Now let us look at the  narmalizer $N=N_G(H\cap K)$ of the intersection in $G.$ It is clear that $HK\subset N$ and therefore we have $|N|$

*

*$\ge27$

*divides $90$

*divisible by $9.$
Hence we have only two possibilities $|N|\in\{45, 90\}.$
If $N=G,$ then $H\cap K$ is normal in $G.$ Otherwise $[G:N]=2$ and then $N$ is normal in $G.$ Therefore we have a contradiction with our assumption and hence $G$ can not be simple.
A: First attempt:
$90=2\cdot 3^2\cdot 5\Longrightarrow\,$ by Sylow theorems, if a group of order 90 is simple then it must have six 5-Sylow subgroups, but then we can make the group act on this sbgps. and thus obtain a homomorphism into $\,S_5\,$ [false, see below], which can't be an injection (why?), contradicting thus the non-existence of normal non-trivial; sbgps. in the group...
Now a little slower: let $\,H\leq G\,,\,[G:H]=n\,$ , and let $\,G^H,$ denote the set of the n left cosets of G in H. Define an action of $\,G\,$ on $\,G^H\,$ by $\,x\cdot gH\to (xg)H\,$ . As any other action, this one defines a homomorphism $\,f:G\to Sym_{G^h}\cong S_n\,$ , by $\,\,f(x)(gH):=(xg)H$.
The nicest part of all this is that $\,\ker f\,$ is the biggest normal sbgp. of $\,G\,$ contained in the sbgp. $H$
Well, the above should suffice to understand the first part. However, see below.
Second attempt:
Of course, m.k.: you are right! I confused the 6 Sylows sbgps. of with the order of each of them , 5...! Of course, this renders my "proof" worthless as 90 is a divisor of 6!
But we can fix this as follows, following the general theory exposed in my first answer: let $\,f:G\to S_6\,$ be the corresponding permutation homomorphism , and let us put $\,N:=\ker f\,$ , so in fact $\,N\leq N_G(P_5)\,,\,\,P_5=\,$a Sylow 5-sbgp.

*

*It can't be $\,N=G\,$ as then $\,P_5\triangleleft G\,$


*If $\,N=1\,$ , then $\,G\lneq S_6\,$ . But any element of $\,G\,$ of order 5 is represented (embedded) within $\,S_6\,$ by a 5-cycle, which is an element of $\,A_6\,$ , and thus $\,A_6\cap G\,$ is a non-trivial proper normal subgroup of $\,G$
Of course, any other case gives us a non-trivial normal proper sbgp. of $\,G$..
Thanks to m.k.
A: Another answer (just rewriting C Monsour's argument for the solvability of $|G|=90$)
We have an injection $f:G\to S_{90}$ and an element of order $2$ in $G$ makes $f(G)\not\subseteq A_{90}$. So in the composition $f:G\to S_{90}\to S_{90}/A_{90}$ it is $imf \not=1$ so $1\not= kerf \lhd G$ (and $kerf\not= G$)
