I've seen this question here before, but I want to know if the following is sufficient:


First note that the product of two normal subgroups $H_1$ and $H_2$ is itself a normal subgroup, and that if $H_1 \cap H_2 = \{e\}$ then $|H_1 H_2| = |H_1||H_2|$. Now suppose we have subgroups $H_1, H_2, \ldots, H_n$, each of which is normal, and such that $\displaystyle \bigcap_{H_j} = \{e\}$. By taking the products one at a time, we obtain that the product $H_1 H_2$ is a normal subgroup of order $|H_1||H_2|$, the product $(H_1 H_2)H_3$ is a normal subgroup of order $|H_1||H_2||H_3| \ldots$, and the product of all the $H_i$, ($i = 1, 2, \ldots, n$) is a normal subgroup whose order is $|H_1||H_2| \cdots |H_n|$.

Now if $G$ is a finite abelian group of order $p_1^{a_1} \cdots p_k^{a_k}$ for distinct primes $p_j$, then the Sylow $p_j$-subgroups $P_1, \ldots, P_k$ have orders $p_1^{a_1} \cdots p_k^{a_k}$, respectively. Note that they are all normal, and that any two distinct Sylow $p_j$-subgroups intersect in the identity. By the argument above, the product $P_1 \cdots P_n$ is a subgroup of of $G$ that has order $p_1^{a_1} \cdots p_k^{a_k} = |G|$, and by the recognition theorem$^\spadesuit$, this product is the same as the direct product, i.e. $G \cong P_1 \times P_2 \times \cdots \times P_k$. Hence $G$ is isomorphic to the direct product of its Sylow subgroups.


Suppose that $x \in P_1 P_2 \cap P_3$. Since $|P_1 P_2|$ and $|P_3|$ are relatively prime, we can write $1 = a|P_1 P_2| + b|P_3|$. Then $$x = x^1 = x^{a|P_1 P_2| + b|P_3|} = x^{a|P_1 P_2|}x^{b|P_3|} = {x^{|P_1 P_2|}}^a{x^{|P_3|}}^b = e.$$

Hence $|x|$ divides 1, so $x = e$. Thus for each $j$, $(P_1 P_2 \cdots P_j) \cap P_{j + 1} = \{e\}$.

$\spadesuit$ Dummit and Foote call the following a "recognition theorem": If $H$ and $K$ are normal subgroups of $G$ and $H \cap K = \{e\}$, then $HK \cong H \times K$.


1 Answer 1


The main idea of the proof is sound, and most of what you've written is good.

A couple of mistakes, though:

  • When you say "Now suppose we have subgroups $H_1,H_2,\ldots,H_n$", and so on, you say nothing about how they intersect, so what follows isn't generally true. Also see the next point.
  • You correctly point out that the Sylow subgroups intersect in the identity, but that's not what you need. You need that $P_1P_2\cdots P_i$ intersects $P_{i+1}$ in the identity for all $i$. That's a stricter requirement, as there are elements in $P_1P_2\cdots P_i$ which aren't in any of the Sylow subgroups.
  • $\begingroup$ Fair points. I've edited my OP to account for your concerns. $\endgroup$
    – Junglemath
    Jul 13, 2020 at 12:06

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