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Let $G$ be a finite group, $N$ a cyclic normal subgroup of $G$, and $H$ any subgroup of $N$. Prove that $H$ is a normal subgroup of $G$.

This proof has errors probably:

Since $N$ is cyclic, let $N$ be generated by $\langle n \rangle$, for some $n \in N$. Also, $N$ is normal in $G$, so we have $gng^{-1} \in N$. Since $H \le G$, closure holds in $H$. So for some $n \in H$ (since $n \in G$), $gng^{-1} \in H$.

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2 Answers 2

up vote 2 down vote accepted

Let $g \in G $.

Observe that $H^{g} \subseteq N^{g} = N$.

N is finite and cyclic, so $H$ is the only subgroup of $N$ with cardinality $|H|$. So $H^{g} = H $ and hence $H$ is normal in $G$.

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Can you clarify "$H^g⊆N^g=N$"? I'm guessing that's an equivalent condition for saying $N$ is normal. – Cookie Feb 25 '14 at 7:13
yes, $N^{g} = N $ for all $g \in G $ because $N$ is normal in $G$ – WLOG Feb 25 '14 at 7:14
@WLOG Can i know what theorem is this called? I don't quite understand why $N^g = N$ – Aha Nov 18 '14 at 2:34
@user2675516: This is a property of cyclic groups; $N^g = N$ because $N$ is normal – WLOG Nov 18 '14 at 22:22

This is a special case of something far more general: a subgroup $K$ of $G$ is called characteristic (and one writes $K$ char $G$) is it is fixed by every automorphism of $G$ ($\forall \alpha \in Aut(G): \alpha(K)=K$).
Because conjugation is an automorphism, every characteristic subgroup is normal, though not every normal subgroup has to be characteristic. Examples of characteristic subgroups include the commutator subgroup and the center of a group. All subgroups of a finite cyclic group are characteristic (because of the unicity of the order of each subgroup).

Now, and this is the general statement, if $K$ char $N \unlhd G$, then $K \unlhd G$. $K$ does not have to be cyclic, an example, let $K=\{(1), (1 2)(34), (1 3)(2 4), (1 4)(2 3)\}$, $N=A_4$ and $G=S_4$. Observe that $A_4'=K$, so $K$ char $A_4$. And hence $K \lhd S_4$.

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It is worth pointing out that $K\unlhd G$ because $G$ acts on $H$ by conjugation and so for every element $g\in G$ there exists an automorphism $\alpha\in\operatorname{Aut}(H)$ such that $g^{-1}hg=\alpha(h)$ for all $h\in H$. That is, $g$ induces some automorphism of $H$. This automorphism fixes $K$ as it is Characteristic, and hence $K^g=\alpha(K)=K$ for all $g\in G$, as required. – user1729 Feb 25 '14 at 14:23

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