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?
We will disprove the claim.
Consider the group $\displaystyle Q_8 = \{\pm1,\pm i,\pm j,\pm k\}$ where $\displaystyle i^2 = j^2 = k^2 = -1, (-1)^2 = 1$. Also we need $\displaystyle ijk = -1$. Using the above relations you can show that $\displaystyle Q_8$ is a group. You can also show it is non-abelian by observing that $\displaystyle ij = -ji$.
Note that $\displaystyle Q_8$ is generated by any two of $\displaystyle i,j,k$.
Now show that if any subgroup $\displaystyle H$ contains $\displaystyle i$ then $\displaystyle \{\pm 1, \pm i\} \subseteq H$. If $\displaystyle H = \{\pm 1, \pm i\}$, then index of $\displaystyle H$ in $\displaystyle Q_8$ is 2 and thus $\displaystyle H$ is normal in $\displaystyle Q_8$. Suppose $\displaystyle \{\pm 1, \pm i\} \subsetneq H$, then one of $\displaystyle j$ or $\displaystyle k$ must belong to $\displaystyle H$. But then $\displaystyle H = Q_8$.
We can proceed similarly for the case where a subgroup $\displaystyle H$ contains $\displaystyle j$ or $\displaystyle k$.
The only remaining case is that $\displaystyle H$ does not contain $\displaystyle i$,$\displaystyle j$ or $\displaystyle k$. In this case $\displaystyle H = \{1,-1\}$, which is the center of $\displaystyle Q_8$ and thus $\displaystyle H$ is normal in $\displaystyle Q_8$