Let $D_8 = <a,b : a^4 = b^2 = 1, bab^{-1} = a^{-1}>$ be the dihedral group.
I'm trying to show that the subgroup generated by $a^2$ is normal. But, isn't $<a^2> ={\{1, a^2}\}$? So the index isn't $2$, so it can't be normal?
What am I missing here? Is my $<a^2>$ right or am I getting this bit wrong? Is it not just powers of $a^2$?
Thank you for any help.
Best Answer
Theorem: Let $G$ be a group and $H\le G$ be a subgroup of $G$. If $[G:H]=2$, Then $H \vartriangleleft G$ is a normal subgroup of $G$.
The converse of the above theorem is not always true. (An obvious example would be $G\vartriangleleft G$.)
Another example is $\langle a^2 \rangle \vartriangleleft D_8$. First, verify that $Z(D_8) = \{1,a^2\}$. And then, in particular, you can conclude that $\langle a^2 \rangle=\{1,a^2\}$ is a normal subgroup of $D_8$.