Semi-direct products: a gentle introduction

In summary, the conversation discusses the concept of semi-direct products in group theory. It explains that semi-direct products are a more complicated version of direct products, where two groups, N and H, interact with each other. The conversation also introduces the concept of an automorphism and shows that in order for the semi-direct product to be a group, there must be a homomorphism from H to the automorphism group of N. The conversation then goes on to define the semi-direct product as well as its underlying set and group operation. It also discusses the special case where the homomorphism from H to the automorphism group of N is trivial, resulting in the direct product. Finally, the conversation proves the three requirements for the semi-direct product
  • #1
Deveno
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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
 
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  • #2
Ok, we've created a new group out of two smaller groups, using what seems to be a really complicated construction. What practical uses could such a beastly thing have? Let's look at an example:

Suppose $N$ is a cyclic group of order $n$. Because cyclic groups are abelian, the inverse map:

$n \mapsto n^{-1}$

is an automorphism:

$(nn')^{-1} = n'^{-1}n^{-1} = n^{-1}n'^{-1}$.

Let's call this map $i$.

It should be clear that $\{1_N,i\}$ forms a subgroup of $\text{Aut}(N)$ of order 2. Since any two groups of order 2 (which are cyclic, of course) are isomorphic, if we take $H = \{e,h\}$ to be a cyclic group of order 2, we have a homomorphism (actually an isomorphism, but that's not important) $H \to \text{Aut}(N)$ given by:

$e \mapsto 1_N$
$h \mapsto i$

So we have all the ingredients we need for a semi-direct product. Clearly, the group we get is of order $|N| \ast |H| = 2n$.

Since $N$ is cyclic, we can write $N = \langle x \rangle$ for some element $x$. We can separate the elements of $N \rtimes_{\theta} H$ into two types:

(a) those of the form $(x^k,e)$
(b) those of the form $(x^k,h)$.

Let's look at the elements of type (a) first. It should be clear these form a subgroup of our semi-direct product, since:

$(x^k,e)\ast(x^m,e) = (x^k\theta_e(x^m),ee) = (x^kx^m,e) = (x^{k+m},e)$

and that this subgroup is isomorphic to $N$ itself. This is indeed true of any general semi-direct product (and it is ALSO true that $\{(e_N,h):h \in H\}$ is a subgroup isomorphic to $H$).

But this subgroup possesses an unexpected property as well: it is NORMAL in the semi-direct product. Behold:

$(x^k,e)\ast(x^m,e)\ast(x^k,e)^{-1} = (x^{k+m},e)\ast(1_N(x^{-k}),e)$

$= (x^{k+m},e)\ast(x^{-k},e) = (x^{k+m-k},e) = (x^m,e)$

so conjugation by an element of type (a) lands us back with an element of type (a), and:

$(x^k,h)\ast(x^m,e)\ast(x^k,h)^{-1} = (x^k i(x^m),h)\ast(i(x^{-k},h^{-1})$

$= (x^kx^{-m},h)\ast(x^k,h^{-1}) = (x^kx^{-m}i(x^k),hh^{-1}) = (x^{k-m-k},e) = (x^{-m},e)$

so conjugation by an element of type (b) also returns an element of type (a).

This is no accident, this always happens in a semi-direct product, the subgroup $N \times \{e_H\}$ is a normal subgroup isomorphic to $N$ (prove this!).

So we know what happens when we multiply two elements of type (a) together, let's look at the other possible types we might get:

1) type (a) times type (b)
2) type (b) times type (a)
3) type (b) times type (b).

$(x^k,e)\ast(x^m,h) = (x^k1_N(x^m),eh) = (x^{k+m},h)$

$(x^k,h)\ast(x^m,e) = (x^k i(x^m),he) = (x^k(x^{-m})h) = (x^{k-m},h)$

Hmm...this is interesting...if we compare (writing 1 for the identity of $N$):

$(x^k,e)\ast(1,h) = (x^k1_N(1),eh) = (x^k,h)$

$(1,h)\ast(x^k,e) = (i(x^k),he) = (x^{-k},h)$

we can see that unless $N$ is of order 2, our semi-direct product is non-abelian.

In fact, if we define the following mapping:

$\phi:N \rtimes_{\theta} H \to D_n$ by:

$\phi (x^k,h^m) = r^ks^m$, where $r$ is a rotation of $2\pi/n$ and $s$ is a reflection about the x-axis, it is not hard to show this is an isomorphism. That is, $D_n$ is a semi-direct product of $\Bbb Z_n$ by $\Bbb Z_2$, and we can write the semi-direct operation ADDITIVELY as:

$(k,m)\ast(k',m') = ((k + (-1)^mk') (\text{mod }n), m+m' (\text{mod } 2))$

The second coordinate tells us if we have a reflection or rotation, and the first coordinate tells us WHICH rotation or reflection we have.

But I digress...let's look at what happens when we multiply two elements of type (b) together:

$(x^k,h) \ast (x^m,h) = (x^k i(x^m),h^2) = (x^{k-m},e)$

We see we get an element of type (a). THIS is why "two reflections composed gives a rotation". How does the element of order 2 in $H$ affect the cyclic group $N$ (that is, what does taking the inverse of a rotation do)? It "flips it over", reversing orientation.

The "jumbling" then, that $H$ does to $N$, is either: nothing (if the second coordinate of the first thing we're multiplying is the identity of $H$), or: reversing the order (if the element of $H$ in the second coordinate of the first term in a product is the non-identity one).

Exercise for the reader:

Realize $S_3$ as the semi-direct product of its two subgroups:

$N = \{e,(1\ 2\ 3), (1\ 3\ 2)\}$ and
$H = \{e, (1\ 2)\}$

using:

$\theta:H \to \text{Aut}(N)$ given by:

$\theta_e = 1_N$

$\theta_{(1\ 2)}(n) = n^{-1}$.

(continued in next post).
 
  • #3
So, so far we've seen we can make a bigger group out of two smaller groups. But there are lots of ways to do this, and moreover, usually the situation is the opposite:

We have some large(-ish) group $G$, which we would like to understand in terms of smaller subgroups of $G$, which we are more comfortable with. This is what the following theorem accomplishes:

Let $G$ be a group with 2 subgroups $H,N$ such that:

1) $N$ is normal in $G$
2) $G = NH$
3) $N \cap H = \{e\}$

Then $G$ is isomorphic to a semi-direct product $N \rtimes_{\theta} H$.

Proof:

The isomorphism we would like to come up with is:

$nh \mapsto (n,h)$

which by (2) and (3) is clearly a bijection. The problem is: how do we come up with $\theta$?

Consider the inner automorphism induced by $h \in H$:

$g \mapsto hgh^{-1}$.

Since $N$ is normal in $G$, this inner automorphism restricted to $N$ yields an automorphism of $N$, that is we define:

$\theta:H \to \text{Aut}(N)$ by:

$\theta_h(n) = hnh^{-1}$.

Now we can define:

$\phi:G \to N \rtimes_{\theta} H$ by:

$\phi(g) = \phi(nh) = (n,h)$

We have already seen that $\phi$ is bijective, it only remains to be seen that $\phi$ is indeed a homomorphism. So let $g = nh, g' = n'h'$.

Then $\phi(gg') = \phi(nhn'h') = \phi(nhn'(h^{-1}h)h') = \phi(n(hn'h^{-1})hh')$

Now, since $N$ is normal, $hn'h^{-1} \in N$, so $nhn'h^{-1} \in N$, and $hh' \in H$, and by (2) and (3) this is the UNIQUE way to write $gg' \in G$ as a product of something in $N$ and something in $H$. Hence:

$\phi(gg') = \phi((nhn'h^{-1})(hh')) = (nhn'h^{-1},hh') = (n\theta_h(n'),hh')$

$= (n,h)\ast(n',h') = \phi(nh)\ast\phi(n'h') = \phi(g)\ast\phi(g')$

and so $\phi$ is a homomorphism.

Exercise left for the reader:

Suppose $G = N \rtimes_{\theta} H$ for two groups $N,H$. Prove:

(a) $N' = N \times \{e_H\}$ is a subgroup of $G$ isomorphic to $N$
(b) $H' = \{e_N\} \times H$ is a subgroup of $G$ isomorphic to $H$
(c) $N'$ is normal in $G$
(d) $N' \cap H' = \{e_G\}$
(e) $G = N'H'$
(f) $(e_N,h)(n,e_H)(e_N,h)^{-1} = (\theta_h(n),e_H)$ (in $G$, the action of $H'$ on $N'$ is just conjugation).

Deduce that $H$ is isomorphic to $G/N$, and thus that there is a surjective homomorphism $G \to H$ with kernel $N$ which is the identity on $H$.
 

FAQ: Semi-direct products: a gentle introduction

What is a semi-direct product?

A semi-direct product is a mathematical concept that combines two groups in a specific way. It is a way to construct new groups from existing ones.

How is a semi-direct product formed?

A semi-direct product is formed by taking two groups, a normal subgroup and a factor group, and combining them in a specific way using a group operation called the semi-direct product operation.

What is the significance of semi-direct products in mathematics?

Semi-direct products are important in mathematics because they allow us to construct new groups from existing ones. They also provide a way to understand the structure of groups and their relationships to each other.

What are some examples of semi-direct products in real life?

Semi-direct products can be seen in many areas of mathematics, physics, and engineering. For example, they are used in the study of crystallography, symmetries of molecules, and the theory of relativity.

What is the difference between a semi-direct product and a direct product?

The main difference between a semi-direct product and a direct product is that in a semi-direct product, the two groups are combined in a non-commutative way, whereas in a direct product, the groups are combined in a commutative way. Additionally, in a semi-direct product, the normal subgroup is not necessarily a subgroup of the factor group.

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