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I want to prove the following:

If every Sylow subgroup of $G$ is a normal subgroup, then $G$ is isomorphic to the product of its Sylow subgroups.

So far, I have gotten to realize that for every prime factor $p_i$ of $|G|$, there is exactly one Sylow subgroup and that these form a partition of $G$. However, I am having trouble finding the isomorphism between $G$ and the product which I will call $G_1 \times \ldots \times G_r$. My idea was just to use the mapping $(g_1, \ldots, g_r) \mapsto g_1\ldots g_r$, which is a bijection, but I don't see whether it is a homomorphism. In general, not all elements of $G_i$ commute with elements of $G_j$, so one cannot simply change the order of the elements in the product.

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1 Answer 1

up vote 3 down vote accepted

Recall the following criterion:

Criterion. A group $G$ is a direct product if and only if it has two normal subgroups $H,K$ such that $H \cap K = \{1\}$ and $G = HK$.

This criterion gives, by induction

Criterion bis. A group $G$ is a $n$-fold product if and only if it has $n$ normal subgroups $H_1,\ldots,H_n$ such that $G = H_1 \ldots H_n$ and $H_i \cap (H_1 \ldots H_{i-1} H_{i+1} \ldots H_n) = \{1\}$.

In you case you have the Sylow subgroups $P_1,\ldots,P_r$ whcih are normal by hypothesis. The equality $G = P_1 \ldots P_r$ follows from cardinality considerations, and $P_i \cap (P_1 \ldots \widehat{P_i} \ldots P_n) = \{1\}$ follows from order considerations.

Indeed, you can change the order in the product! In fact, let $g \in G_i$, $h \in G_j$. Then $g h g^{-1} h^{-1} \in G_j \cap G_i$. But $G_j$ is a $p_j$-group and $G_i$ is a $p_i$-group! If an element lies in their intersection, its order must divide both $p_i^n$ and $p_j^m$, hence it is $1$.

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Thank you. My book has a different treatment of products and deals normal subgroups after products, so I have never seen the criterion. The right to left implication is proven, however, in a slightly different manner (requiring all elements of H to commute with all elements of K, but this follows from the fact that both are normal subgroups, as you mentioned in the last line of your answer). The other implication is not proven, however, and I am trying to see why it also holds. –  Peter Dec 4 '12 at 20:46
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The other direction is the easier one: if $G \simeq G_1 \times \ldots \times G_n$ then you have normal subgroups $H_i := \{1\} \times \ldots \times G_i \times \ldots \times \{1\}$, and these subgroups do satisfy the requirements of the criterion. –  Mauro Porta Dec 4 '12 at 21:06

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