0

My question: For odd prime $p$, what is the easiest group-theoretic way of showing that $\mathbb{Z}_p^2\rtimes C_2$ has a normal subgroup of size $p$?

Attempt: If $\mathbb{Z}_p^2$ has a characteristic subgroup of size $p$, then we are done (since if $A$ is char in $B$ and $B$ is normal in $C$, then $A$ is normal in $C$). But this doesn't help since $\mathbb{Z}_p$ is not a characteristic subgroup of $\mathbb{Z}_p^2$.

  • The answer already gave you a good way to do the actual problem. On the other question: No, any group which is the product of isomorphic simple groups is characteristically simple (i.e. has no proper nontrivial characteristic subgroups). – Tobias Kildetoft Apr 24 '17 at 08:18
  • @ Tobias, thanks for letting me know that my hint (attempt) doesn't help then since $Z_p$ is not a characteristic subgroup of $Z_p^2$. So what's d best group theoretic way of showing that $Z_p^2\rtimes C_2$ has a normal subgroup isomorphic to $Z_p$ for prime $p\geq 3$? –  Apr 24 '17 at 08:21
  • Exactly (and if we picked another prime instead of $2$, it could happen that there was no subgroup like the one you are looking for here). – Tobias Kildetoft Apr 24 '17 at 08:23

1 Answers1

1

Think Linear Algebra. A linear transformation $T$ of order $2$ on a finite-dimensional vector space over $\mathbb{F}_p$ is diagonalisable, and so there are fixed subspaces.

ancient mathematician
  • 14,102
  • 2
  • 16
  • 31