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I don't believe the following statement is true in general

A normal subgroup of prime order is central.

But then what is wrong with my logic here?

Suppose $N \triangleleft G$ with $|N| = p$ prime. Since $N$ is closed under conjugation by elements of $G$, it is a union of conjugacy classes. Since it contains the identity, its contribution to the class equation of $G$ is some partition of $p$ containing at least one $1$. The number of ones appearing in this partition is the size of the intersection of the center of $G$ and $N$, which must be a subgroup of $N$. The only possibility is $1+\ldots+1$ ($p$ ones), meaning all five elements of $N$ are in the center of $G$. Therefore $N$ is in the center of $G$.

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  • $\begingroup$ A simple counterexample is $S_3$ which has trivial centre. It has a single normal subgroup of prime order, $A_3$. What's the size of the intersection here? You'd need something like $p$ is the smallest prime dividing $\#G$ for this to work in general. $\endgroup$ Mar 4 '15 at 22:58
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Why can't the intersection be just the identity element? That's a subgroup of $N$ which is also a subgroup of the centre of $G$.

Some similar arguments use counting and divisibility by $p$ to show that there is more than one element in the centre. For example the argument that the centre of a group of order $p^r$ for prime $p$ and positive integer $r$ is non-trivial - and has at least $p$ elements. It looks as though you are trying to emulate this proof, but that doesn't work here, because there is no such restriction on the order of $G$ in your question.

Another way of seeing this is that the automorphism group of $N$ has order $p-1$ (and is cyclic) and there can be elements $g\in G$ which give non-trivial automorphisms of $N$ by conjugation. If the order of $g$ had to be a power of $p$ this would not be the case, but this restriction is not present in this case.

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  • $\begingroup$ Note, a specific example is the non-abelian group of order $6$, which has a normal subgroup of order $3$ which is not central. The non-abelian group of order $21$ has a normal subgroup of order $7$ which is not central. Note that if $G/Z$ is cyclic then $G=Z$, so any nonabelian group of order $pq$ will have a cyclic subgroup of prime order which is not in the centre. $\endgroup$ Mar 4 '15 at 23:00

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