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Given an elementary abelian $p$-group $G$, it's well known that it can be seen as a vector space over $\mathbb{Z}_p $.

But, does someone have an idea about possible sources where I can find proofs that use this fact? Examples: linear independence of elements in a group $G $ modulo $\phi(G) $ ($\phi(G)$ is the Frattini group of $G$ ), etc. ...

Does someone have an idea?


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coincidentally, see… – Yemon Choi Aug 23 '12 at 12:30
Not very deep, but the proof that a group that has only the identity automorphism must be $C_2$ uses this fact (together with the Axiom of Choice to construct of a basis). – Arturo Magidin Aug 23 '12 at 15:50
On another tack, in trying to characterize the capable $p$-groups of exponent $p$ and class $2$, I ended up looking at certain morphisms between the commutator subgroups of the relatively free groups of exponent $p$ and class 2 and those of exponent $p$ and class $3$. I then interpreted them as linear maps between vector spaces, and worked in the linear algebra setting instead (because I found it more easy to think about). Some of that stuff is in the arXiv: – Arturo Magidin Aug 23 '12 at 15:53
Great ! THanks ! Have you got any reference for the automorphism proof? – Jason Mraz Aug 24 '12 at 7:17
@Evgeny: Here's one:; related to this question is Asaf Karagila's proof that there are models of ZF in which there is a nontrivial vector space over $GF(2)$ of dimension greater than 1 that has no nontrivial automorphism: – Arturo Magidin Aug 24 '12 at 15:24
up vote 3 down vote accepted

One example is in the $\sigma$-reduction of a finite group. Denote by $\sigma(|G|)$ the number of prime divisors of $|G|$, and let $\sigma(G)$ denote the maximum number of prime divisors of the order of an element of a group $G$. We call a group $H$ $\sigma$-reduced if every prime $p$ appears in at most one chief factor of $H$. It can be shown that every group $G$ contains a subgroup $H$ so that $\sigma(|G|)=\sigma(|H|)$. (Reference: Huppert's Character Theory of Finite Groups, chapter 16. In fact, this chapter also has a few other applications of this concept to Frobenius groups.)

There is an ongoing investigation of a bound on $\sigma(|G|)$ by $\sigma(G)$ for finite solvable groups. (The conjecture is that $\sigma(|G|)\leq 3\sigma(G)$.) After fixing a $\sigma(G)$, it suffices to consider the $\sigma$-reduction of $G$ to obtain the bound on $\sigma(|G|)$ while enjoying the properties of $\sigma$-reduced groups. In particular, Sylow subgroups of the $\sigma$-reduction are elementary abelian, so interplay between the bases of representations over vector spaces of $\mathbb{Z}_p$ for $p\mid |G|$ play a crucial role. You can see some of this happening in Thomas Keller's earlier papers Solvable groups with at most four prime divisors in the element orders and Solvable groups with a small number of prime divisors in the element orders.

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@Alexander: Thanks ! It indeed seems like a reasonable direction to look at. The problem is that it also seems like a subject that require some advance knowledge... I'm looking for something a bit more elementary... Have you got any other suggestions? Thanks a lot again ! – Jason Mraz Aug 23 '12 at 9:41
Like I said, Ch. 16 in Huppert has a few other theorems of this type. Yemon Choi's comment above is also a good place to look. – Alexander Gruber Aug 23 '12 at 15:50

Proofs of the following simple facts use the vector space structure of $G/\Phi(G)$:

Let $G$ be a $p$-group with $\dim_{\mathbb{F}_p} G/\Phi(G)=n$. Then:

  1. $n$ is the minimal number of generators of $G$.
  2. The number of maximal subgroups of $G$ is $\dfrac{p^n-1}{p-1}.$

For the first statement see Robinson: A Course in the Theory of Groups, 5.3.2 (Burnside's Basis Theorem). For the second, let $\kappa: G \to G/\Phi(G)$ and note that there is a map

$$F: Hom_{\mathbb{F}_p}(G/\Phi(G),\mathbb{F}_p) \setminus \lbrace 0 \rbrace \to \lbrace M \le G \text{ maximal }\rbrace=: \mathcal{M},\;f \mapsto \ker(f \circ \kappa).$$ We have $F(f_1)=F(f_2)$ iff $f_2 = \alpha f_1$ for some $\alpha \in \mathbb{F}_p^\times$. Futhermore, $F$ is surjective, since $M\le G$ maximal induces $G/\Phi(G) \to G/M\cong \mathbb{F}_p$. Thus $|\mathcal{M}|=(\dim G/\Phi(G)-1)/(p-1)$.

As an example let $G=D_8$ be the dihedral group of order 8. From $G/\Phi(G)=\mathbb{Z}/2 \times \mathbb{Z}/2$ we conclude that $D_8$ has three maximal subgroups (in fact, a cyclic subgroup and two elementary abelian subgroups each of order 4).

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Thanks a lot !!! – Jason Mraz Aug 24 '12 at 7:21

Linear algebra can be applied to exploit the structure of extraspecial groups. That's a p-group $G$ where the commutator subgroup equals the center $C$ and has order p. $V:= G/C$ is an elementary abelian p-group, because $G/C$ is abelian and $$[g,h^p]=gh^pg^{-1}h^{-p}=[g,h]h[g,h^{p-1}]h^{-1}=[g,h][g,h^{p-1}]=\cdots=[g,h]^p=1.$$
It's easy to see that $\varphi: V \times V \to C\cong \mathbb{F}_p,\; (\bar{x},\bar{y}) \mapsto [x,y]$ defines a bilinear form that is skew-symmetric because $[x,y]^{-1}=(xyx^{-1}y^{-1})^{-1}=[y,x]$ and non-degenerate (for, if $[x,y]=1$ for all $y$, then $x \in C$ and $\bar{x}=0$ in $V$).

Now, by linear algebra, each vector space that admits a non-degenerate skew-symmetric bilinear form can be decomposed into orthogonal subspaces $V=\oplus_{i=1}^n V_i$ where $V_i$ is two-dimensional with basis $\bar{x}_i,\bar{y}_i$ such that $\varphi(\bar{x}_i,\bar{y}_j)= \delta(i,j)$ and $\varphi(\bar{x}_i,\bar{x}_j)=\varphi(\bar{y}_i,\bar{y}_j)=0$ for all $i,j$.

In particular $\dim V=2n$ what shows $|G|=|C| \cdot |V|= p^{2n+1}$ and the formulas for $\varphi$ give the presentation $$\begin{array}{ll}G = \langle x_1,y_1,...,x_n,y_n,c \mid & c^p=[x_i,c]=[y_i,c]=1,[x_i,y_i]=c, \newline & x_i^p=c^{k_i}, y_i^p=c^{l_i},[x_i,x_j]=[y_i,y_j]=[x_i,y_j]=1\rangle \end{array}$$ where $i\neq j$ run through $1,...,n$ and $0 \le k_i,l_i < p$.

Putting some more work into it, one can show $k_i=l_i=1$ if p is odd while in case p=2 the two cases $(k_1,l_1)=(1,0),(1,1)$, $k_i=l_i=0$ if $i> 1$ occur.

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I wanted to use $\delta_{i,j}$ for the Kronecker delta (instead of $\delta(i,j)$ in the text) but mysteriously that doesn't work. Any idea ? – Todd Leason Aug 24 '12 at 10:26

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