I was looking at the group of order 12, and it was said that when there are three 2-Sylow subgroups, the intersection of those(order 2) must be characteristics.

I showed that the intersection of Sylow 2-subgroup must be equal to the kernel of conjugation action under G on 2-Sylow subgroups(3), but I couldn't figure out why this kernel should be characteristics. There are three subgroups of order 4, but how can we show from this to say that there is only one subgroup of order 2, which is kernel?


This is true in general.

Let $G$ be a finite group, and let $p$ be a prime dividing the order of $G$. If $P$ is a Sylow $p$-subgroup and $\phi$ an automorphism of $G$, then $\phi(P)$ must also be a Sylow $p$-subgroup (why?).

Now, let $S$ be the intersection of all Sylow $p$-subgroups in $G$. To be characteristic, we just need to show that $\phi(S)\subseteq S$. Since $\phi$ just permutes the Sylow $p$-subgroups, the map $\phi$ must preserve their intersection as well — which is exactly $S$. Thus $S$ is characteristic.

In similar spirit, try to prove the following fact:

Let $G$ be a finitely generated group, and $n$ some integer. Prove that for every subgroup $H$ in $G$ with index $n$, there exists a finite-index normal (or characteristic!) subgroup of $G$ inside $H$.

Hint 1:

Whenever $G$ is finitely generated, there are only finitely many subgroups of index $n$ for any $n$.

Hint 2:

The intersection of finite index subgroups is a finite index subgroup.

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  • $\begingroup$ I can clearly understand. Thank you so much! $\endgroup$ – Min Seok Song Dec 13 '19 at 16:58

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