Can comeone help me? How do I show that every group of order 90 is not simple?
$90=2\cdot 3^2\cdot 5\Longrightarrow\,$ 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.
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.