# Normal subgroups

• Aug 24th 2010, 10:55 PM
shos
[SOLVED] Normal subgroups
we know that if a group is abelian, all subgroups of it are normal. however - can you prove or disprove that if all subgroups of a group are normal, it is abelian?
• Aug 24th 2010, 11:10 PM
Isomorphism
We will disprove the claim.

Consider the group $Q_8 = \{\pm1,\pm i,\pm j,\pm k\}$ where $i^2 = j^2 = k^2 = -1, (-1)^2 = 1$. Also we need $ijk = -1$. Using the above relations you can show that $Q_8$ is a group. You can also show it is non-abelian by observing that $ij = -ji$.
Note that $Q_8$ is generated by any two of $i,j,k$.

Now show that if any subgroup $H$ contains $i$ then $\{\pm 1, \pm i\} \subseteq H$. If $H = \{\pm 1, \pm i\}$, then index of $H$ in $Q_8$ is 2 and thus $H$ is normal in $Q_8$. Suppose $\{\pm 1, \pm i\} \subsetneq H$, then one of $j$ or $k$ must belong to $H$. But then $H = Q_8$.

We can proceed similarly for the case where a subgroup $H$ contains $j$ or $k$.
The only remaining case is that $H$ does not contain $i$, $j$ or $k$. In this case $H = \{1,-1\}$, which is the center of $Q_8$ and thus $H$ is normal in $Q_8$