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Let $G$ be a group of order $p^2$ ($p$ being a prime) and $N$ be a normal subgroup of $G$ such that $O(N)=p.$ Without using the fact any group of order $p^2$ is abelian or sylow theorem how to show that $N\subset Z(G).$

Added: The problem is from the the article Cayley's theorem of Herstein where the concept of conjugancy classes is yet to be introduced. So please don't use the concept of conjugancy classes.

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If you're familiar with automorphisms: for any normal subgroup $N\trianglelefteq G$ there is a map $\varphi:G\to{\rm Aut}(N)$ sending $g\in G$ to the conjugation automorphism $\alpha_g(x)=gxg^{-1}$. The automorphism group of $N$, which is of size $p$ hence cyclic, is $U(p)=({\bf Z}/p{\bf Z})^\times$ of order $p-1$. Since the image of $G$ under $\varphi$ divides $\#G$ it must be among $\{1,p,p^2\}$, but it must also divide $p-1$ by Lagrange's theorem, which means the image must be the trivial subgroup. So conjugation on $N$ by the rest of $G$ is trivial, i.e. $gxg^{-1}=x \forall x\in N,g\in G$. –  anon Apr 13 '13 at 14:53
    
I'm loving your restrictive, non-Sylow questions, by the way. –  Alexander Gruber Apr 13 '13 at 16:55
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4 Answers

I will try to stay within the restrictions you place. If the group is cyclic of order $p^{2},$ there is nothing to do. Otherwise, every element of $G \backslash N$ (and there are some!) has order $p.$ Let $M$ be a subgroup of $G$ of order $p$ different from $N.$ Then Cayley's theorem gives ( via theaction of $G$ on the right cosets of $M$ by right translation) a homomorphism $\phi: G \to S_{p}$ whose kernel is the largest normal subgroup contained in $M.$ But $|G| = p^{2}$ does not divide $|S_{p}|,$ so ${\rm ker \phi} \neq 1,$ and $M \lhd P.$ Now for $m \in M $ and $n \in N,$ we have $m^{-1}n^{-1}mn =m^{-1}(n^{-1}mn) = (m^{-1}n^{-1}m)n \in M \cap N =1.$ Thus $mn = nm$ Since $G =MN,$ we see that $N \subseteq Z(G).$

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Ah haha, $p^2$ does not divide $S_p$ - that's slick! +1 –  Alexander Gruber Apr 13 '13 at 16:43
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If $N$ is the only subgroup of order $p$, then any other subgroup must have order $1$ or $p^2$, so $G$ must be cyclic because it has a unique subgroup of every order $d$ for each divisor $d$ of $|G|$. In this case, $Z(G)=G$ so we are done. Suppose then that there is another subgroup of order $p$ called $M$. Then $N\cap M=1$. Since $|G|=p^2=|N||M|$ we then have that $G$ is an internal direct product of $N$ and $M$, so $G\cong \mathbb{Z}_p\oplus \mathbb{Z}_p$, which is abelian, and once again we are finished.

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Hint 1. Every normal subgroup is a disjoint union of conjugacy classes.

Hint 2. The size of a conjugacy class divides the order of the group.

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Look at the group operation of $G$ on $N$ by conjugation. The neutral element is a fixed point. Since $p$ is prime, the remaining $p-1$ elements of $N$ must be fixed points, too (since the size of each orbit divides the group order $p^2$). So all elements of $N$ are fixed under conjugation and therefore in the center.

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