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Semi-direct products are often confusing to the budding group theorist, with good reason: they are a bit more complicated than the more transparent direct products.
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:
http://mathhelpboards.com/commentary-threads-53/commentary-gentle-introduction-semidirect-products-7854-post35916.html#post35916
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:
http://mathhelpboards.com/commentary-threads-53/commentary-gentle-introduction-semidirect-products-7854-post35916.html#post35916