Why does all order $n$ subgroups of an order $2n$ group have to be normal?

Asked Sep 28 '18 at 16:24

Active Sep 28 '18 at 16:24

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I see it for $D_3$ and $D_4$ but I would like to prove it for order $n$.

I have these definitions:

A normal subgroup $H$ of a group $G$ is a subgroup that consists of complete conjugacy classes of $G$.

A conjugacy class $C$ is those set of elements which commute, ie $$g_2 = g g_1g^{-1}; \forall g\in G; g_1,g_2 \in C $$

So an order $2n$ group, $G_{2n}$ can have order $n$ subgroups, $SG_n=C_1 \cap C_2 ... C_k$

Why all order $n$ subgroups of $G$ have to be normal?

asked Sep 28 '18 at 16:24

zabop

1,011

2

Because the group index $[H:G] = 2$. – GNUSupporter 8964民主女神地下教會 Sep 28 '18 at 16:25
2

$n$ of the group elements belong to the subgroup, and the other $n$ elements belong to the other left coset. The same other $n$ elements belong to the other right coset too. – Theo Bendit Sep 28 '18 at 16:26

Why does all order $n$ subgroups of an order $2n$ group have to be normal?

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