Show that a group of order 66 has a normal subgroup of order 33.

Question

This question is somewhat similar to: A group of order $66$ has an element of order $33$.

However, I do not understand how I would show that the subgroup of order 33 is normal. So far I have that there is a unique Sylow 11-subgroup, but I don't know how to continue.

Any hints are appreciated.

Every subgroup of index $2$ is normal - see for example here. — Dietrich Burde, Dec 29 '18 at 23:45
Related; https://math.stackexchange.com/questions/84632/subgroup-of-index-2-is-normal — Servaes, Dec 29 '18 at 23:46
I would say that your question is fully covered by this more general result. I won't vote to close right away, because there is scope for disagreement. I have the dupehammer on [tag:group-theory], so my vote would be immediately binding. — Jyrki Lahtonen, Dec 30 '18 at 14:17

score 3 · Accepted Answer · answered Dec 29 '18 at 23:55

Let $P$ be the Sylow $11$-subgroup you've shown exists. Since you've shown $P$ is unique, we have that $P$ is normal. Now if we let $Q$ be a Sylow $3$-subgroup, then $PQ=QP$ by normality of $P$, and thus $PQ$ is a subgroup of $G$. But $PQ$ has order $33$ (hence index $2$) so $PQ$ is normal in $G$.

score 1 · Answer 2 · edited Dec 30 '18 at 00:08

1

We know by the linked answer that there exists an element of order $33$. Let this element be $x$. Hence, $\langle x\rangle$ has order $33$. We know that all subgroups with index 2 in a group are normal.

edited Dec 30 '18 at 00:08

MJD

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answered Dec 29 '18 at 23:52

Rushabh Mehta

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Show that a group of order 66 has a normal subgroup of order 33.

2 Answers2