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- Feb 15, 2012

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First off, let's start with a (my apologies) formal definition:

We need 2 groups to start with: we shall call these groups (for reasons that hopefully will be clearer later on) $N$ and $H$.

The goal is to build a "bigger group" out of $N$ and $H$, but one in which $N$ and $H$ "interact"...for the time being, I will just say that $H$ "jumbles" $N$ before we "put them together". The language here is to make you think that $H$ "permutes" $N$ in some way, and that is exactly what we are going to do.

Since $N$ is a group, we don't just want our induced mapping $N \to N$ to be a bijection (an "ordinary" permutation), we would like it to be a homomorphism. But what is a bijective homomorphism from $N \to N$ called? An automorphism of $N$.

We would like this automorphism to at least preserve "something" of the group character of $H$. How do we do that? We insist that we have a homomorphism of $H$ into $\text{Aut}(N)$.

So in addition to our two groups, we insist we have one more ingredient: a homomophism $\theta: H \to \text{Aut}(N)$. Now that we have all our ingredients, we can cook them up.

We say we have a semi-direct product of $N$ by $H$ over $\theta$, written:

$N \rtimes_{\theta} H$

if we have a homomorphism $\theta: H \to \text{Aut}(N)$.

Well, that's all very well and good, but we haven't given even the bare basics of what a group should have:

1) An underlying set

2) A group operation

We should fix this, hmm?

(1) is the easy part: we will take as our underlying set, $N \times H$ (a pair: the first element of the pair lies in $N$, the second one lies in $H$). That seems straight-forward enough, right? This makes it clear that if our groups $N,H$ are finite:

$|N \rtimes_{\theta} H| = |N| \ast |H|$.

(2) is where things get a bit sticky: in order to get something we haven't seen before, $\theta$ is going to "gum up the works". Before we continue, a slight notational detour:

Normally, we would write $\theta(h)$ for the image of an element $h$ under $\theta$. However, $\theta(h)$ is a FUNCTION $N \to N$, and it's cumbersome (and a bit confusing) to write:

$\theta(h)(n)$

for the image of an element $n$ under the function which is $\theta(h)$. So we will instead write:

$\theta(h) = \theta_h \in \text{Aut}(N)$

so that we can write $\theta_h(n)$ for the image of $n$ under the automorphism $\theta_h$.

Now for the hard part (again, my apologies):

For our product, we set:

$(n,h)\ast(n',h') = (n\theta_h(n'),hh')$

As you can see, the second coordinate is not any different than the direct product, it's just the first one that is a little strange. Let's consider a very important special case, first:

It might be, that no matter which element of $H$ we pick, the resulting "jumble" of $N$ ISN'T one, that is, every element of $H$ induces the identity permutation of $N$. This is what we get if the homomorphism $H \to \text{Aut}(N)$ is trivial: that is:

$\theta_h = 1_N$ for all $h \in H$. In that case, our product becomes:

$(n,h) \ast (n',h') = (n\theta_h(n'),hh') = (n1_N(n'),hh') = (nn',hh')$

(since $1_N(n) = n$ for all $n \in N$), which is just the "ordinary" direct product.

So we see that the direct product is just a "special case" of the semi-direct product.

Now, although I *have* displayed a FUNCTION:

$(N \rtimes_{\theta} H) \times (N \rtimes_{\theta} H) \to (N \rtimes_{\theta} H)$

so we know we have a binary operation, I have NOT yet shown that this is a GROUP operation, which means we need to show (3) things:

(1) $\ast$ is associative

(2) $\ast$ has an identity in $n \times H$

(3) every $(n,h)$ has an inverse element in $N \times H$ under $\ast$.

Let's show (2) first, because it's easy:

I claim $(e_N,e_H)$ is an identity for $\ast$. Let's check this:

$(n,h)\ast(e_N,e_H) = (n\theta_h(e_N),he_H) = (n\theta_h(e_N),h)$.

Now since $\theta_h$ is an automorphism of $N$ (and thus a homomorphism), it maps the identity of $N$ to itself (it HAS to). Thus $\theta_h(e_N) = e_N$, so we have:

$(n,h)\ast(e_N,e_H) = (n\theta_h(e_N),h) = (ne_N,h) = (n,h)$. Halfway there. Now we check the other side:

$(e_N,e_H)\ast(n,h) = (e_N\theta_{e_H}(n),e_Hh) = (e_N\theta_{e_H}(n),h)$.

But now, because $\theta$ is a homomorphsim, it has to map $e_H$ to the identity of $\text{Aut}(N)$, and the identity of the automorphism group is the identity map on $N$. Thus $\theta_{e_H}(n) = (1_N)(n) = n$, so:

$(e_N,e_H)\ast(n,h) = (e_N\theta_{e_H}(n),h) = (e_Nn,h) = (n,h)$.

This proves (2).

(1) is going to be nasty....it's really a straight-forward application of the definition, but it's going to get messy. We need to show that:

$(n_1,h_1)\ast[(n_2,h_2)\ast(n_3,h_3)] = [(n_1,h_1)\ast(n_2,h_2)]\ast(n_3,h_3)$

Working on the left side, we get:

$(n_1,h_1)\ast[(n_2,h_2)\ast(n_3,h_3)] = (n_1h_1)\ast(n_2\theta_{h_2}(n_3),h_2h_3)$

$= (n_1(\theta_{h_1}(n_2\theta_{h_2}(n_3)),h_1(h_2h_3))$

Wow, that first coordinate really looks ugly. However, since $\theta_{h_1}$ is a homomorphism, we have:

$\theta_{h_1}(n_2\theta_{h_2}(n_3)) = \theta_{h_1}(n_2)\theta_{h_1}(\theta_{h_2}(n_3)) = \theta_{h_1}(n_2)((\theta_{h_1}\circ \theta_{h_2})(n_3))$

Since $\theta$ is a homomorphism of $H$ into $\text{Aut}(N)$, we have:

$\theta_{h_1} \circ \theta_{h_2} = \theta(h_1h_2) = \theta_{h_1h_2}$.

So we can re-write our LHS as:

$(n_1(\theta_{h_1}(n_2)\theta_{h_1h_2}(n_3)),h_1(h_2h_3))$

and now using associativity in $N$ and $H$ we see this equals:

$((n_1\theta_{h_1}(n_2))\theta_{h_1h_2}(n_3),(h_1h_2),h_3)$ (**).

Now let's work on our right-hand side:

$[(n_1,h_1)\ast(n_2,h_2)]\ast(n_3,h_3) = (n_1\theta_{h_1}(n_2),h_1h_2)\ast(n_3,h_3)$

$= ((n_1\theta_{h_1}(n_2))\theta_{h_1h_2}(n_3),(h_1h_2)h_3)$

which is just what we obtained in (**), so the two sides are equal. Fortunately, we only have to prove this once (whew!).

For our last trick, we go on to exhibit an inverse. Once again, we will use a special property of $\theta$: since it is a homomorphism, we have that:

$\theta_{h^{-1}} = \theta_h^{-1}$

This means, in particular, that:

$\theta_h \circ \theta_{h^{-1}} = \theta_{e_H} = 1_N$

Since we have a two-sided identity, it suffices to show that we have a one-sided inverse (although you may check we have a two-sided inverse for yourselves, as an exercise):

$(n,h)\ast(\theta_{h^{-1}}(n^{-1}),h^{-1}) = (n\theta_h(\theta_{h^{-1}}(n^{-1})),hh^{-1}$

$ = (n((\theta_h \circ \theta_{h^{-1}})(n^{-1})),e_H) = (n,1_N(n^{-1}),e_H) = (nn^{-1},e_H) = (e_N,e_H)$

Thus $(\theta_{h^{-1}}(n^{-1},h^{-1})$ is an inverse for $(n,h)$.

Thus $(N \times H,\ast)$ truly is a group.

(continued in next post).

Questions and comments should be posted here:

Commentary for "Semi-direct products: a gentle introduction".