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I'm trying to work out this group theory proof:

Suppose that $G$ is a group have that has a normal subgroup $H$ such that $H$ is isomorphic to $D_3$. Prove that exists a subgroup $K$ of $G$ such that $G$ is isomorphic to $H\oplus K$.

Here is what I was thinking:

I was thinking $K$ could be the centralizer of $H$ in $G$. Because then we know that $K$ is a normal subgroup of $G$ since $H$ is normal. Then I was thinking that there would be a relationship between a group $G$ and the direct product of two normal subgroups, however I'm not really sure how to build on that.

Also I was wondering how we can use the fact that $H$ is isomorphic to $D_3$, because I can't really understand how that would help.

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Let $H \cong D_3$ and $K = C_G(H)$ (centralizer of $H$ in $G)$. Then $G/K \cong A \le Aut(D_3)$, $K \trianglelefteq G$.

Here you have to prove that $|Aut(D_3)| = |D_3|$ (this can be done "manually", by checking all the options where automorphism can send the generators of $D_3$).

Since $Z(D_3) = \{e\}$, every automorphism of $D_3$ is inner. $H \le G$, therefore $G/K \cong Aut(D_3)$, with $|K| = \dfrac{|G|}{|H|}$.

Since $Z(H) = \{e\} =>$ $K \cap H = \{e\}$.

$|HK| = \dfrac{|H|\times|K|}{|H \cap K|} = |G|,$ hence $HK = G$.

Then you can use the proof outlined here:

Direct product of two normal subgroups

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