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Given two groups $A, B$, we can construct direct product $A \times B$ whose elements are of the form $(a, b), a \in A, b\in B$. If $A, B$ are subgroups of a group $G$ and $A \cap B =\{1\}$, then we can construct semidirect product $A \rtimes B$ whose elements are of the form $ab, a\in A, b\in B$. In this case, is the semidirect product $A \rtimes B$ the same as direct product $A \times B$? Is the order of $AB$ equal to $|A||B|$? For example, is the order of $(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ equal to $2^n * n!$? Here $S_n$ is the symmetrical group of order $n$. Thank you very much.

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Heuristically, the difference is this: If $A, B\unlhd G$, $G=AB$ and $A\cap B=1$ then $G\cong A\times B$, while if $A\unlhd G$, $B\leq G$, $G=AB$ and $A\cap B=1$ then $G\cong A\rtimes B$. So, in the former both subgroups are normal while in the latter we only require one to be. Note that direct products are semidirect products. If we assume that neither group need be normal then we get the Zappa-Szep product of $A$ and $B$, $G\cong A\bowtie B$. – user1729 Jul 26 '12 at 19:52
up vote 2 down vote accepted

In your second sentence, $A$ is required to be a normal subgroup. The semi direct product $A \rtimes B$ will always have the same order as the direct product $A \times B$, but in a direct product $B$ will be normal too.

Take $A$ to be order 3, $B$ to be order 2, and consider two non-isomorphic groups of order 6. The cyclic group of order 6 will contain $A \times B$, but the symmetric group of degree 3 will contain $A \rtimes B$ that is not a direct product.

Weyl group example

The hyperoctahedral group $B_n$ is defined to be a semi-direct product $(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ and can be explicitly represented as the group of all integer matrices such that (1) all non-zero entries are ±1, and (2) each row and each column has exactly one nonzero entry.

The subgroup $A=(\mathbb{Z}/2\mathbb{Z})^n$ corresponds to the diagonal matrices with diagonal entries all being ±1, so $2^n$ different possibilities. This is a normal subgroup of the hyperoctahedral group.

The subgroup $B$ of permutation matrices, the group of all integer matrices such that (1) all non-zero entries are 1, and (2) each row and each column has exactly one nonzero entry, is also a subgroup of the hyperoctahedral group. However, it is not normal.

You can see this form the presentation of the hyperoctahedral group on its Coxeter generators:

$$\left\langle s_1, s_2, \ldots, s_n : s_1^2 = s_i^2 = 1, (s_1 s_2)^4 = (s_i s_{i+1})^3 = (s_i s_j)^2 = 1 \mid 2 \leq i \leq n, i+2 \leq j, j \leq n \right\rangle$$

where the generator $s_1$ is the diagonal matrix with a single $-1$ in its top left entry, and 1s on the rest of the diagonal, and the generators $s_i$ are the permutation matrix formed from the identity by switching both the $i-1$st and $i$th rows, and the corresponding columns.

If this was a direct product, then we would need $n$ generators for $A$. In the semi-direct product, the elements of $B$, especially $s_i$ for $2 \leq i \leq n$, move the first generator $s_1$ of $A$, to the other $n$ generators needed. In the direct product $(s_1 s_2)^2$ would be the identity, but in the hyperoctahedral group, it is the matrix whose first row has a $-1$ in the second position, whose second row has a 1 in the first position, and in all other rows, the 1 is in the same place as for an identity matrix.

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Let me add that there is an abstract definition of the semi-direct product.

Let $A$ and $B$ be groups and let $\varphi: B \to Aut(A)$ be a morphism, i.e. an action of $B$ on $A$. Given that you can define the semi-direct product $A \rtimes_\varphi B$ as the set $A\times B$ with group law $$ (a,b) * (a',b') = (a\varphi(b)(a'),bb') $$ (In practice, $\varphi$ is often omitted in the notation $A \rtimes_\varphi B$ but the group law strongly depends on it: it tells you how $A$ and $B$ mix in the product).

Now it is clear from the definition that

  • $|A \rtimes_\varphi B| = |A||B|$ since the underlying set is the product set
  • If $\varphi$ is the trivial morphism then $A \rtimes_\varphi B$ is the usual direct product since $\varphi(b)(a') = a'$ in this case.

In fact you can prove that as subgroups of $A \rtimes_\varphi B$, the two sub-groups $A$ and $B$ commute (meaning that $ab=ba$ forall $a \in A$, $b\in B$) if and only if $\varphi$ is trival (so that the semi-direct product is actually a direct product).

PS: To compare with other definitions of the semi-direct product, simply identify $A$ with $\{(a,1)~|~a\in A\}$ and $B$ with $\{(1,b)~|~b\in B\}$. Also note that if $A$ and $B$ are subgroups of $G$ with $A$ normal, then you can take the action by conjugation as $\varphi$: $\varphi(b)(a) = bab^{-1}$.

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This construction is sometimes (often?) called the "external" semidirect product, while the $G=AB$, $A\cap B=1$, $A\unlhd G$ is called the "internal" semidirect product. – user1729 Jul 26 '12 at 21:01

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