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Are there finite groups $G$ and $H$ such that:

  1. $n:=|G|=|H|$.
  2. $G$ is abelian.
  3. $H$ is nonabelian.
  4. for every $d\mid n$, $G$ and $H$ have the same number of elements of order $d$. ?

I know two finite abelian groups with the same numbers of elements of each order; are isomorphic. And there are nonabelian groups with the same property which are not isomorphic.

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I misunderstand a point. Let $G$ be cyclic and $H$ is an non-abelian. And so we accept 1,2,3. $G$ has an element of order $n$. we assume 4 is included in our assumption, what would happen if $H$ has an element of order $n$? –  B. S. Jan 27 '13 at 17:09
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I think, @Babak, that G must then be abelian, but not cyclic. –  amWhy Jan 27 '13 at 17:10
    
Yes $G$ can not be cyclic, since If $G$ is cyclic,then has an element of order $n$, but $H$ has no an element of order $n$. –  maryam Jan 27 '13 at 17:12
    
@amWhy: It seems to me that $G$ should be just abelian not cyclic. Or, at least, some restrictions needed for $d$. I hope, I am on a right way. –  B. S. Jan 27 '13 at 17:12
    
if G is cyclic, H is also cyclic and abelian. There's only one cyclic group of any order up to isomorphism. –  user59671 Jan 27 '13 at 17:38
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4 Answers

up vote 8 down vote accepted

This is similar to the answer by Andreas.

Let $p > 2$ be prime, $G$ be elementary abelian of order $p^3$, $H$ nonabelian of order $p^3$ such that $x^p = 1$ for each $x \in H$. That is, $G \cong \mathbb{Z}_p \times \mathbb{Z}_p \times\mathbb{Z}_p$ and $H$ will be the Heisenberg group

$$H \cong \left\{ \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : a, b, c \in \mathbb{Z}_p \right\}$$

In both groups there are $p^3 - 1$ elements of order $p$.

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Similar, but yours is simpler. I wonder why I thought of the case of exponent $p^2$ instead of the simpler case of exponent $p$. –  Andreas Caranti Jan 27 '13 at 17:13
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Yes, for any odd prime $p$ there is

  • one (isomorphism type of) abelian group $G$ of order $p^3$ and exponent $p^2$, and
  • one nonabelian group $H$ of order $p^3$ and exponent $p^2$,

and they have the same number of elements of each possible order, that is,

  • 1 element of order 1,
  • $p^2 - 1$ elements of order $p$, and
  • $p^3 - p^2$ elements of order $p^2$.

$G$ is the product of a cyclic group of order $p^2$ with a (cyclic) group of order $p$. As to $H$, it has a presentation $\langle a, b : a^{p^2} = b^{p} = 1, [a, b] = a^p \rangle$.

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So an example can be found among groups of order 8. thanks. –  user59671 Jan 27 '13 at 17:03
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@CutieKrait odd prime. –  Andres Caicedo Jan 27 '13 at 17:04
    
@CutieKrait No, this works for an odd prime only. –  Andreas Caranti Jan 27 '13 at 17:04
    
.............................. of order 27. thanks. –  user59671 Jan 27 '13 at 17:07
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The extraspecial $p$-group ($p$ an odd prime) $$E_n = \langle x_1,y_1,\ldots,x_n,y_n,z \mid x_i^p=y_i^p=1, z=[x_i,y_i] \textrm{ central }, [x_i,x_j]=[x_i,y_j]=1\rangle$$ ($i,j=1,\ldots,n,\;i\neq j$) has order $p^{2n+1}$ and exponent $p$ (and is non-abelian). Hence it has the same number of elements in each order as $C_p^{2n+1}=C_p \times \cdots \times C_p$.

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kind of generalization. –  user59671 Jan 27 '13 at 17:52
    
Yes, the Heisenberg group in m.k.'s answer is just $E_1$. –  tj_ Jan 27 '13 at 18:29
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Generalising a bit the example of m.k. Let $p$ be an odd prime and $F$ any field of characteristic $p$. Then $GL(n,F)$ does not have any elements of order $p^2$ as long as $n\leq p$: the minimal polynomial of such an element must be a power of $X-1$, but $X^p-1=(X-1)^p$ is divisible by $(X-1)^i$ for all $i\leq n$ in this case, so that elements with minimal polynomial $(X-1)^i$ still have order $p$.

Therefore the group $U$ of upper unitriangular matrices in such $GL(n,F)$, as well as any subgroup of $U$, have the property that all its non-identity elements have order $p$, a property they share with the elementary Abelian $p$-group of the same order, while the group $U$ in not Abelian when $n\geq3$ (this is in fact the only reason $p\neq2$ is relevant).

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