I would like to improve my proof of the following result:

If $H$ is a finite, elementary abelian $p$-group, then $\Phi(H) = 1$.

Here, $\Phi(H)$ is the Frattini subgroup, defined as the intersection of all maximal subgroups of $H$. An elementary abelian $p$-group is an abelian group with the property that $x^p = 1$ for all $x \in H$.

I proved this by choosing an arbitrary nonidentity element $x \in H$ and showing that there exists a maximal subgroup $M < H$ which does not contain $x$.

My proof explicitly constructs $M$. I wonder if there is a cleaner, more elegant way to establish the existence of $M$ without resorting to a construction.

My proof: Let $x \in H$ be a nonidentity element. Define $N_1 = \langle x \rangle$. Then $|N_1| = p$. If $N_1 < H$, choose $x_2 \in H \setminus N_1$, and define $N_2 = \langle x_2\rangle$. Then $N_1 N_2$ is a subgroup, with $N_1 \cap N_2 = 1$, so $N_1 N_2$ is a direct product with order $p^2$.

If $N_1 N_2 < H$, we continue in the same way. By step $k$ we start with a direct product $N_1 N_2 \cdots N_{k-1}$ with order $p^{k-1}$. If this subgroup is proper, choose $x_k \in H \setminus N_1 N_2 \cdots N_{k-1}$ and define $N_k = \langle x_k \rangle$. Then $N_1 N_2 \cdots N_k$ is a direct product with order $p^k$.

Since $H$ is finite, this process eventually terminates and we end up with $H = N_1 N_2 \cdots N_m$, where $N_1 = \langle x \rangle$. Then $M = N_2 \cdots N_m$ is a maximal subgroup of $H$ which does not contain $x$:

  • Maximality is clear because the index of $N_2\cdots N_m$ is $p$.
  • If $x \in N_2 \cdots N_m$ then we can write $x$ in two different ways in $N_1 N_2 \cdots N_m$, namely $(x)(1)\cdots(1) = (1)(n_2)\cdots(n_m)$. The uniqueness of representation in a direct product then forces $x = 1$.

Context: This question addresses a remark made by Isaacs in Finite Group Theory, following problem 1.D.8:

If $G$ is a finite $p$-group, then $\Phi(G)$ is the unique normal subgroup of $G$ minimal with the property that the factor group is elementary abelian.

I proved this by establishing the following results:

  1. If $H$ is a finite, elementary abelian $p$-group, then $\Phi(H) = 1$.
  2. If $\theta : G \to H$ is a surjective homomorphism, then $\theta(\Phi(G)) \leq \Phi(H)$.
  3. Therefore, if $K \lhd G$ and $G/K$ is elementary abelian, then $\Phi(G) \leq K$.

1 Answer 1


Let $H$ be an elementary abelian $p$ group. Then we know that $H\cong Z_p\times Z_p...\times Z_p$.

Set $$M_1=1\times Z_p\times ... Z_p$$ $$M_2=Z_p\times 1\times ..\times Z_p$$ $$M_n=Z_p\times ...\ Z_p\times 1$$

Notice that $M_i$ are maximal in $G$ and $$1=\cap_{i=1}^n M_i\geq \Phi(H)$$.

Note: Notice that $A\times B\cap C\times D=(A\cap B)\times (C\cap D)$.

  • $\begingroup$ Thanks, this is a much nicer way! $\endgroup$
    – user169852
    Dec 15, 2015 at 18:23
  • $\begingroup$ you are welcome. $\endgroup$
    – mesel
    Dec 15, 2015 at 18:24

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