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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.