Prove a cyclic subgroup is normal? I've got a question that asks me to find a cyclic subgroup of $D_{12}$ of order 6.  $D_{12} = <r, s | r^{12} = s^2 = e, sr = r^{11}s>$
My answer to this is :
{$e, r^2, r^4, r^6, r^8, r^{10}$} or {$e, r^2s, r^4, r^6s, r^8, r^{10}s$}.
The question I have for you is that I'm then asked to prove that whichever subgroup I choose is a normal subgroup.  I would normally say that it is a normal subgroup, because it is cyclic and therefore Abelian and normal too.  Also, it must be a subgroup of $D_{12}$ because it is generated by an element of $D_{12}$.  However, it is a 4 mark question, so am I missing something?
Thanks
 A: Knowing $r,s$ generates $G$, we only need to check $rHr^{-1}=sHs^{-1}=H$ to show a subgroup $H$ is normal in $G$.
This is clear for your subgroups. But the second subgroup seems not cyclic.
A: i believe that a theorem along the lines "let $p$ be the smallest prime dividing the order of the group $G$.  If $H\leq G$ is a subgroup of index $p$, then $H$ is normal in $G$" will help you find the answer
A: First, note  that the subgroup $\langle r \rangle = \{e, r, r^2, \ldots, r^{11}\}$ has index $2$ in $D_{12}$ and is therefore normal in $D_{12}$. Moreover, $\langle r \rangle$ is cyclic, so it contains exactly one subgroup of each possible order. This means that every subgroup of $\langle r \rangle$ is characteristic in $\langle r \rangle$.
There is a useful general theorem which is easy to prove: if $H \lhd G$ and $K$ is a characteristic subgroup of $H$, then $K \lhd G$. Applying this result with $K = \langle r^2 \rangle = \{e, r^2, r^4, \ldots, r^{10}\}$ and $H = \langle r \rangle$ and $G = D_{12}$, we can conclude immediately that $K \lhd G$.
In fact, by the same reasoning, every subgroup of $\langle r\rangle$ is normal in $G$.

For completeness, here is a proof of the theorem used above. Suppose that $H \lhd G$ and $K$ is characteristic in $H$. The latter means exactly that any automorphism of $H$ fixes $K$, in other words if $\phi \in \operatorname{Aut}(H)$ then $\phi(K) = K$. Let $g$ be any element in $G$, and let $\phi_g$ denote conjugation by $g$. Then $\phi_g$ is an automorphism on $G$. Moreover, since $H \lhd G$, we see that $\phi_g$ fixes $H$ (i.e. $\phi_g(H) = H$), so $\phi_g|_H$ (the restriction of $\phi_g$ to $H$) is an automorphism of $H$ and therefore fixes $K$. In other words, $\phi_g(K) = K$. But $\phi_g(K) = gKg^{-1}$, so by definition this means that $K \lhd G$.
