# A finite abelian group is isomorphic to the direct product of its Sylow subgroups

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

Attempt:

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.

EDIT:

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 $$HK \cong H \times K$$.

• 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.