Showing a Group $G$ is not Simple

Question

Let $G$ be a finite group of order $pq$, where $p,q$ are distinct prime numbers. Show that $G$ is not simple.

Here is my attempt:

$|G|=pq$. If $G$ is not simple, then it has non-trivial subgroups, i.e., subgroups other than the identity and itself.

Let $H$ be a subgroup of order $p$. By Lagrange's Theorem, $|H| \mid |G| \Rightarrow p \mid pq$. Similarly, let $K$ be a subgroup of order $q$ so that $|K| \mid |G| \Rightarrow q \mid pq$. Hence, $G$ is not simple.

I was hoping somebody could verify my proof and point out any errors. As always, any help or advice is greatly appreciated.

Careful: it must be "If $;G;$ is not simple then it has non-trivial normal subgroups..." . This (but not only: you also have a logic error) renders your whole proof wrong. — Timbuc, Nov 24 '14 at 15:17
Note that by Cauchy's theorem $G$ must have a subgroup of order $p$. The trick is showing that the subgroup is normal. — Ben Grossmann, Nov 24 '14 at 15:25
@DietrichBurde, thanks for linking the two questions. After taking a look at it, they do seem to be very similar. — Jamil_V, Nov 24 '14 at 15:37

score 1 · Accepted Answer · answered Nov 24 '14 at 15:28

1

Hints:

Suppose WLOG that $\;p>q\;$ , then

== How many Sylow $\;p$- subgroups are there in the group?

== If and only if what a Sylow subgroup is normal?

answered Nov 24 '14 at 15:28

Timbuc

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Showing a Group $G$ is not Simple

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