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show that in a group G of order p^2 any normal subgroup of order p must lie in the center of G.

user110666
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  • What have you tried? suppose $N\unlhd G$ with $|N|=p$ and $N\cap Z(G)= {e}$ then.. what could possibly go wrong? –  Dec 11 '13 at 11:37
  • I think there is some link between right or left cosets , commutativity of Z(G) – user110666 Dec 11 '13 at 11:43
  • Note that we do not actually need the order to be $p^2$ for this. Any power of $p$ will do. I know two ways to do this, one using the normalizer/centralizer theorem, and the other by (slightly annoying) calculations. Have you heard of the normalizer/centralizer theorem? – Tobias Kildetoft Dec 11 '13 at 11:45
  • NO.. i donot no that theorem... – user110666 Dec 11 '13 at 11:46
  • this question is from the book Topics in Algebra by I.N.Herstein – user110666 Dec 11 '13 at 11:47
  • If $x$ is an element in the given subgroup and $g$ is an element from the group, then you know that $gxg^{-1}$ is some power of $x$. You need to show that this power must be $1$ mod $p$. Note what hapens if you instead conjugate by some power of $g$, and use that $g^{p^k}$ is the identity for some $k$. – Tobias Kildetoft Dec 11 '13 at 11:59
  • The title and body don't match. The title says the group has order $p$, the body, $p^2$. Can you edit please for clarity? – Gerry Myerson Dec 11 '13 at 12:01

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Prove the following :

  1. If $|G| = p^n$, then $Z(G) \neq \{e\}$ (Use the class equation)

  2. If $G/Z(G)$ is cyclic, then $G$ is abelian.

  3. If $|G| = p^2$, then $G$ is abelian. Hence, $Z(G) = G$

Conclude whatever you want :)

Prahlad Vaidyanathan
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  • he said he do not know class equation... I do not think this would help him any how.. –  Dec 11 '13 at 12:17