How can I show that if some group $G$ has only one subgroup $K$ of order $n$, then $K$ is a normal subgroup? Would that mean that it only has one subgroup total? If so then I guess that makes sense.
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1Hint: conjugation preserves order. – Alex Youcis Oct 02 '13 at 07:08
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Isn't that the definition? – Oct 02 '13 at 07:12
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1@Tony: No, that's not the definition, it is the observation you can use to see that conjugation gives another group of order $n$, hence you can apply your hypothesis that there is only one such group. "Would that mean..." No, there can be other groups with different orders. For example, $\mathbb Z_{32}$ has only one subgroup of order $2$, but it has a few other subgroups with different orders. – Jonas Meyer Oct 02 '13 at 07:14
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To flesh out Alex Youcis's suggestion, we recall that a subgroup $K < G$ is normal iff for every $g \in G$, $gKg^{-1} = K$. But always we have that $gKg^{-1}$ is a subgroup of order $n$ (since conjugation is an automorphism). So it is another subgroup of order $n$. But there is only one, $K$, so $gKg^{-1}=K$ for all $g$.
AlexM
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