$(a)$ Let $T_1$, $T_2$, $\cdots$, $T_k$ be non-abelian finite simple groups. How many normal subgroups does the direct product $T_1 \times T_2 \times \cdots \times T_k$ have?

$(b)$ Let $G$ denote the elementary abelian group of order $p^n$, i.e. $$G \cong \underbrace{(\mathbb{Z}/p\mathbb{Z}) \times \cdots \times (\mathbb{Z}/p\mathbb{Z})}_{n\text{ times}}$$ How many subgroups of order $p^k$ does $G$ have?

For part $(b)$, I think we might be able to count the number of elements of order $p^k$, and each of these generate a subgroup. Then divide by $p^k - 1$ to correct for overcounting. But I'm not sure how to count the number of elements like this.

And I have no idea how to start part $(a)$. Any help would be much obliged!

  • $\begingroup$ Hint for part $(b)$: $G$ is an $n$-dimensional vector space over the field with $p$ elements. $\endgroup$ – Sameer Kailasa Jan 1 '15 at 23:28

$(a)$ Let $G = T_1 \times \cdots \times T_k$ and let $N \unlhd G$. Denote $\pi_i : G \to T_i$ the projection homomorphism. Observe that $\pi_i (N) \unlhd T_i$, so $\pi_i (N) = (e)$ or $\pi_i (N) = T_i$ for each $1\le i \le k$.

Assume $\pi_{i} (N) = T_i$ for $i\in S$ and $\pi_{i}(N) = (e)$ for $i\not\in S$. This means that $N \le \prod_{i\in S} T_i$.

Now, let $G'$ denote the commutator subgroup of $G$. Since the $T_i$ are non-abelian simple, $\pi_i (G') = T_i$ for all $1\le i \le k$.

Choose $x\in T_j$, where $j\in S$. By the previous observation, $x = aba^{-1}b^{-1}$ for some $a$, $b\in T_j$. Since $j\in S$, there is an element $(t_1, \cdots,b, \cdots, t_k)\in N$, where $b$ is in the $j$th slot. Normality of $N$ implies $$(t_1, \cdots, a, \cdots, t_k) (t_1, \cdots,b, \cdots, t_k) (t_1, \cdots, a, \cdots, t_k)^{-1} = (t_1, \cdots, aba^{-1}, \cdots, t_k) \in N$$ Then, as $(t_1, \cdots,b, \cdots, t_k)^{-1} \in N$, it follows $(e, \cdots, aba^{-1}b^{-1}, \cdots, e) = (e, \cdots, x, \cdots, e) \in N$. Therefore, we also have the reverse inclusion: $ \prod_{i\in S} T_i \le N$.

Hence, $N$ is the direct product of some $T_i$'s. We conclude that $G$ has exactly $2^k$ normal subgroups, one for each subset of $\{1, \cdots, k\}$.

$(b)$ Notice that $G$ is an $n$-dimensional vector space over $\mathbb{F}_p$. It is not hard to see that subgroups of order $p^k$ correspond with subspaces of dimension $k$. Thus, we must count the number of subspaces of dimension $k$.

First, choose a nonzero vector $v_1\in G$. There are $p^n -1$ choices for $v_1$. Then, choose a vector $v_2$ not in the span of $v_1$; since there are $p$ vectors in the span of $v_1$, there are $p^n - p$ choices for $v_2$. Continuing in this fashion, there are $$(p^n - 1)(p^n - p)\cdots (p^n - p^{k-1})$$ ways to choose an ordered $k$-tuple of linearly independent vectors in $G$. The span of each $k$-tuple gives a subspace of dimension $k$. But each subspace has many choices of basis, and we must divide by the number of bases to correct for overcounting. By a similar argument to above, there are $$(p^k -1)(p^k - p) \cdots (p^k - p^{k-1})$$ choices of basis for each $k$-dimensional subspace. Thus, the number of $k$-dimensional subspaces (read: subgroups of order $p^k$) is $$\frac{(p^n - 1)(p^n - p)\cdots (p^n - p^{k-1})}{(p^k -1)(p^k - p) \cdots (p^k - p^{k-1})} = \frac{(p^n - 1)(p^{n-1} - 1) \cdots (p^{n-k+1} - 1)}{(p^k - 1)(p^{k-1} - 1) \cdots (p -1)}$$

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    $\begingroup$ Very helpful and clear, thank you! $\endgroup$ – Spencer Hyman Jan 2 '15 at 0:06

For part (a):The trivial group and the group itself are the only normal subgroup of any simple groups; therefore, for the direct product each tuple is going to be either trivial or the entire group so you will have $2^k$ many normal subgroups.

  • $\begingroup$ But not every subgroup of $T_1 \times \cdots \times T_k$ must be a direct product of $T_i$. $\endgroup$ – Spencer Hyman Jan 1 '15 at 23:30

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