In the question titled Prove that if |G|=132 then G cannot be simple, it is shown a group of order $132$ cannot be simple. My summary of the proof:
- assume group is simple
- deduce the number of Sylow $11$– and $3$-subgroups ($12$ & $4$).
- deduece how many elements we have left to put into subgroups (we have used up $(11-1)\cdot12+(3-1)\cdot4=128)$, so we have $132-128=4$ left
I understand the proof up to this point. But then it continues:
So then $n_3=4$. There are only $4$ remaining elements, which must
comprise a Sylow $2$-subgroup which is unique and thus normal but this
is a contradiction.
Here I don't follow. If we have assumed group is simple, that means there are no unique Sylow $p$-groups. If we did, that Sylow $p$-group would be normal, making $G$ not simple. I base this on the lemma I read about in these notes (source code – see also the bottom of this question).
If I have $4$ elements left to deal with (including the identity), why cannot these be $a, b, c, I$ where $a^2 = b^2 = c^2 = I$?
Ie:
$$\mathrm{Syl}_2(G)=\{\{a,I\},\{b,I\},\{c,I\}\}$$
making $n_2=|\mathrm{Syl}_2(G)|=3$, which is allowed.
So no contradiction is found after assuming $G$ is simple. It is a problem though, since both the referenced thread, and other sources I saw claim $G$ is not simple.
What am I getting wrong?
Lemma: If $n_p = 1$, then the Sylow $p$-subgroup is normal in $G$.
Proof: Let $P$ be the unique Sylow $p$-subgroup, and let $g \in G$, and consider $g^{-1}Pg$. Since this is isomorphic to $P$, we must have $|g^{-1} Pg| = p^a$, i.e. it is also a Sylow $p$-subgroup. Since there is only one, we must have $P = g^{-1}Pg$. So $P$ is normal.
Best Answer
There's only $4$ remaining elements, and if there's more than one Sylow $2$ there would need to be more than $4$.
(Thus the Sylow $2$ subgroup is unique, hence normal, hence $G$ isn't simple.)
What you're getting wrong is that the Sylow $2$ has order $4$, not $2$.