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.

  • 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
up vote 1 down vote accepted

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 $D_8$.

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