Let $G$ be a finite group such that whose center is of odd order and the order of inner automorphisms groups is 6. Then prove that $G$ has a non abelian normal subgroup of order 6.

Thank you


Since $G/Z$ cannot be cyclic, we have $G/Z \cong S_3$, where $Z = Z(G)$ had odd order. Let $a$ and $b$ be inverse images in $G$ of elements of $S_3$ of orders $3$ an $2$. Then $|b| = 2m$ with $m$ odd and by replacing $b$ by $b^m$ we can assume that $|b|=2$.

Now $b^{-1}ab = a^{-1}z$ for some $z \in Z$, so $a^3 = b^{-1}a^3b = a^{-3}z^3$.

Let $c = a^{-1}b^{-1}ab = a^{-2}z$. Then $c^3=a^{-6}z = 1$, and $b^{-1}cb = (a^{-1}z)^{-2}z = a^2z^{-1} = c^{-1}$, so $\langle c,b \rangle \cong S_3$ and (since $Z$ is central), $G = Z \times \langle c,b \rangle$, and $\langle c,b \rangle$ is the required normal subgroup.


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