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Let $G$ be a $p$-group of class 2 and $\exp(\operatorname{Inn}(G))=p^c$. Then prove $\frac{G} {Z (G)}$ has the form $C_{ p^c}\times C_{ p^c}\times C$ for some (possibly trivial) abelian $p$-group $C$.

$C_n$ denotes the cyclic group of order $n$. Thank you

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2 Answers

up vote 6 down vote accepted

OK, here is a solution! Since $G$ has class 2, $G/Z(G)$ is abelian. Let me write $C(i)$ for a cyclic group of order $p^i$. Then $G/Z(G) \cong C(c) \times C(c_1) \times \cdots C(c_k)$ with all $c_i \le c$. We are asked to proved that there exists $c_i$ with $c_i=c$. So assume not. Then all $c_i < c$.

Let $x,y_1,y_2,\ldots,y_k$ be inverse images in $G$ of generators of the cyclic direct factors of $G/Z(G)$. I claim that $x^{p^{c-1}} \in Z(G)$, which is a contradiction, because $xZ(G)$ is assumed to have order $p^c$ in $G/Z(G)$. Clearly $[x,x^{p^{c-1}}]=1$. Since each $y_i^{p^{c-1}} \in Z(G)$, and the commutator map is bilinear in nilpotent groups of class 2, we have $[x^{p^{c-1}},y_i] = [x,y_i^{p^{c-1}}] = 1$ and hence $x^{p^{c-1}}$ centralizes all generators of $G/Z(G)$, so it lies in $Z(G)$ as claimed, giving a contradiction.

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Hints...and disclaimer:

$$(1)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;G/Z(G)\cong\operatorname{Inn}(G)$$

$$(2)\;\;\;\;\;\;\;\;G/Z(G)\,\,\,\text{cannot be cyclic non-trivial}$$

Disclaimer: The fact that at least two factors $\,C_{p^c}\,$ appear in the decomposition of $\,G/Z(G)\,$ follows now immediately, but the fact that there are at most two of them escapes me right now, though I think the class two thingy has something seriously important to do with this...perhaps later.

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I am not quite seeing why at least two factors of the form have to be there, since there are plenty of non-cyclic groups of the right exponent with just one such factor. Ohh, and note that the question does not require that $C$ has no further factors like the first ones. –  Tobias Kildetoft Jan 9 '13 at 16:33
4  
I agree it does not seem completely obvious that there are at least two factors of that form! To see that, you need to use the fact that, for class two groups, the commutator map is bilinear. So if there was only a single $C_{p^c}$ factor generated mod $Z(G)$ by $x \in G$, then $[x^{p^{c-1}},y]$ would be trivial for all $y \in G$, giving $x^{p^{c-1}} \in Z(G)$, contradiction. –  Derek Holt Jan 9 '13 at 17:54
    
Very nice point, @DerekHolt. Thanks. Yet I don't understand why it must precisely be $\,[x^{p^{c-1}},y]\,$ trivial: wouldn't we just have that $\,[x^k,y]\,$ be trivial, for some $\,k=p^d\,\,,\,1\neq d\mid c\,$? Or it has to be $\,d=c-1\,$ precisely? –  DonAntonio Jan 9 '13 at 20:58
    
If there is only a single $C_{p^c}$ factor, then $G$ is generated mod $Z(G)$ by $x$, and elements $y_1,\ldots,y_k$, where the orders of the $y_i$ mod $Z(G)$ are $p^{c_i}$ with $c_i < c$. Then $x^{p^{c-1}}$ commutes with $x$, and $[x^{p^{c-1}},y_i] = [x,y_i^{p^{c-1}}]= 1$ (since $c_i\le c-1$), so $x^{p^{c-1}}$ commutes with all generators mod $Z(G)$ and hence it is central. –  Derek Holt Jan 10 '13 at 9:16
    
@DerekHolt You just answered the question really nicely, I wish you would write it down as an answer :) (Also the fact that class 2 implies inn G abelian was not mentioned yet.) –  Myself Jan 10 '13 at 18:17
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