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 $HK \cong H \times K$.


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 '20 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.