# 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?

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This looks like a homework problem, so please tag it as such. And what have you tried. What sort of theorems do you know? Just Sylow, or do you know more advanced stuff? – Tobias Kildetoft May 18 '12 at 13:20

### 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.

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

2. 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.

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How do you get a homomorphism into $S_5$? Do you mean $S_6$ since there are six Sylow subgroups? – Mikko Korhonen May 18 '12 at 15:30
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. i) It can't be $\,N=G\,$ as then $\,P_5\triangleleft G\,$ ii) If $\,N=1\,$ , then $\,G\lneq S_6\,$ . – DonAntonio May 18 '12 at 16:16
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 $\,G$.. Thanx to m.k. – DonAntonio May 18 '12 at 16:16
@DonAntonio: First: this is something you should bring up in meta, not here. Second, you can add material, you don't have to replace it. Third, everyone can see the history of anything that is changed after five minutes (it is only changes done during that "some few minutes" that are completely invisible; if your proposal had merit, it would make far more sense to do it the other way: forbid changes in the first few minutes, because those are invisible to everyone, and only allow them after, because those are reviewable by everyone). – Arturo Magidin May 18 '12 at 16:58
@Don, it is, of course, perfectly fine to just leave hints for the OP. Many of us do that. But then we let everybody know that this is just a hint/sketch/whatever. And, yes, we are sticklers for having answers largely free of errors, so we fix rather than post anew. Also, if you wanted your comment to reach me, you can use the at-ping mechanism. Then I get notified by the system! Learn to use it! Comes in really handy. It will even autocomplete my name for you, in case you have trouble spelling. And last: welcome to the site, you'll learn the ropes soon enough! – Jyrki Lahtonen May 18 '12 at 17:11