Let $G$ be a non-abelian group, $|G|=6$. Then there exists $a\in G$ such that $o(a)=2$. Is it true that $\langle a\rangle$ is not normal in $G$?

Question

Let $G$ be a non-abelian group of order 6. Then by Cauchy's Theorem, there exists $a\in G$ such that $o(a)=2$. Is it true that $\langle a\rangle$ is not normal in $G$?

My work:

Assume that $\langle a\rangle$ is normal. Then $\langle a\rangle\subseteq Z(G)$.

Edit- We cannot use the fact that $G$ is isomorphic to $D_6$.

Answer 1

Without Sylow theorems and dihedral groups:

Let $g$ be any element of $G$.Any conjugate $gag^{-1}$ of $a$ has order $2$ so that, if $\langle a \rangle$ were normal, $gag^{-1}=a$, i. e. $ga=ag\;$ and the group would be abelian.

Some details:

Any other element has order $2$ or $3$ by Lagrange's theorem. There can be no other element $b$ with order$2$, otherwise $\langle a,b\rangle$ would be a subgroup of order $4$. Hence any element $b\neq e,a\;$ has order $3$ and $G$ would be the commutative (cyclic) group generated by $a$ and $b$.

Answer 2

Any non-abelian group of order 6 is $D_{6}$, the Dihedral group acting on the three-cycle. The subgroup of order 2 in $D_{6}$ is the subgroup of reflections, which is not normal in $D_6$. Let the subgroup of rotations act on the reflections by conjugation and you will see that the reflections are not preserved.

Answer 3

If $N$ and $M$ are normal subgroups of $G$ with $NM=G$ and $N\cap M=\{e\}$ then $G\cong N\times M$.

Let $G$ be a group of order $6$, notice that $v_3=1$, so the $3$-sylow subgroup is normal, if the $2$-sylow subgroup where also normal then $G\cong P\times Q$, where $P$ is the $2$-sylow subgroup and $Q$ is the $3$-sylow subgroup, so $G$ would be a direct product of abelian groups, and hence abelian.

Answer 4

You're nearly done. By Cauchy's theorem there is also an element of order $3$. Since the element of order $2$ is in the center, you can conclude that the group must be abelian since it is generated by two commuting elements of orders $2$ and $3$. By assumption the group is not abelian, so it follows that a subgroup of order $2$ is not normal.

Answer 5

You have already proved that $\langle a \rangle$ normal in $G$ implies $\langle a \rangle\le Z(G)$; but then (Lagrange and nonabelian $G$), $|Z(G)|=2$ and hence $G/Z(G)\cong C_3$: contradiction.