# Normal subgroup generated by $D_8$

Let $$D_8 = $$ be the dihedral group.

I'm trying to show that the subgroup generated by $$a^2$$ is normal. But, isn't $$ ={\{1, a^2}\}$$? So the index isn't $$2$$, so it can't be normal?

What am I missing here? Is my $$$$ right or am I getting this bit wrong? Is it not just powers of $$a^2$$?

Thank you for any help.

• Index 2 implies normal, but the converse doesn't hold in general. Note that $\langle a \rangle$ has index 2 and hence is normal. Also, $\langle a \rangle$ is cyclic, so it has a unique (hence characteristic) subgroup of each possible order. In particular, $\langle a^2 \rangle$ is the unique order-2 subgroup of $\langle a \rangle$, hence $\langle a^2 \rangle$ is characteristic in $\langle a \rangle$. Then use the useful general fact that if $N$ is normal in $G$ and $H$ is characteristic in $N$, then $H$ is normal in $G$. – Bungo Nov 8 at 20:14

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 $$D_8$$.