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My abstract algebra class introduced me to direct products, not semidirect products. I became interested in semidirect products when confronted with the following homework problem:

Define the quaternion group $Q_8 = \{1, i, j, i \mid i^2 = j^2 = k^2 = ijk = -1\}$. Show that the set $\mathbb{H} = \{a + bi + cj + dk \mid a, b, c, d \in \mathbb{R}\}$ is a group under multiplication.

Rather than checking all the properties, I wanted to use my intuition that $\mathbb{H}$ was a group because it "combined" $Q_8$ and $\mathbb{R}^4$, which were both groups. That is, any element of $\mathbb{H}$ can be uniquely specified by choosing an element of $Q_8$ and an element of $\mathbb{R}^4$. So I wrote down

$\mathbb{H}$ is a group because it is isomorphic to $Q_8 \times \mathbb{R}^4$

which I now see is clearly wrong.

Now, the Internet tells me (I think) that what I was actually thinking about was the semidirect product, not the direct product. So that's my first question:

  • Is $\mathbb{H}$ a semidirect product of $Q_8$ and $\mathbb{R}^4$? Which acts on which (which belongs on the left side of $\rtimes$?)

My second question: I'm forming the intuition that, given normal subgroup $N$ and subgroup $H$, one forms $G$ by:

  1. Defining an element of $g$ as a unique combination of an element of $N$ and an element of $H$.
  2. Giving rules for how multiplication between different elements $h_1n_1$ and $h_2n_2$ will work. If you choose:

$$(h_1n_1)(h_2n_2) = (h_1h_2)(n_1n_2)$$

then it's just a direct product, but you might choose anything, such as:

$$(h_1n_1)(h_2n_2) = (h_1h_2^{-1})(n_1n_2)$$

How inaccurate is this vague idea? In particular, is the above a valid choice for a way to multiply elements of $N \rtimes H$? What restrictions are there on these choices?

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  • $\begingroup$ As a general rule, when you have a semidirect product, the action is usually obvious. If you have to ask "which acts on which", it is probably not a semidirect product of those two groups. $\endgroup$ – Ted Apr 4 '16 at 5:29
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    $\begingroup$ also, H is not a group, only H* (H excluding 0) is a group. $\endgroup$ – Ted Apr 4 '16 at 5:29
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    $\begingroup$ What is true is that the quarternions are a quotient of the real group algebra of $Q_8$. But this construction isn't directly related to direct or semidirect products. It's not true that a quarternion is specified by an element of $Q_8$ and an element of $\mathbb{R}^4$, and even if it were that still wouldn't be enough because the multiplication in the quarternions has nothing directly to do with the group structure on $\mathbb{R}^4$. $\endgroup$ – Qiaochu Yuan Apr 4 '16 at 5:34
  • $\begingroup$ @QiaochuYuan: D'oh -- you're certainly right. So a large part of my question is invalid. So now I'm curious about "quotient of the real group algebra of $Q_8$" -- is that the name of the process you use to construct $\mathbb{H}$ out of $Q_8$? $\endgroup$ – Eli Rose Apr 5 '16 at 13:56
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As Ted says, when you have a semidirect product the action is usually natural. If it is not, then you shouldn't try to force it, because this means that there is no semi-direct product involve.

However when you have a group $G$ and try to write it as a semi-direct product, you have to find a group $N$ normal in $G$ and a complement $K$ to it (to finally end up with $G=N\rtimes K$).

In the present case, here are some questions you should answer first :

  • What is the law group on $\mathbb{H}$?

There is essentially only one on it, it is the addition. For this addition structure $\mathbb{H}$ is abelian and isomorphic to $\mathbb{R}^4$. I assume that you rather mean $\mathbb{H}^*$ with the multiplication.

  • The group $Q_8$ is a subgroup of $\mathbb{H}^*$ , is it normal?

No.

  • Is there a group in $\mathbb{H}^*$ isomorphic to $\mathbb{R}^4$?

No.

So, we are far from having a semi-direct product structure in this case. I would like to end up with an explanation about the intuition you should have when talking about semi-direct product.

If in a group $G$ you have two subgroup $N,K$ such that any element $g$ of $G$ is written as a product $g=nk$ with $n\in N$ and $k\in K$, this means that the application :

$$\psi:N\times K\rightarrow G$$ $$(n,k)\mapsto nk$$

is surjective. If you add a second hypothesis that $N\cap K$ is trivial and $N$ normal in $G$ then, this implies that $\psi$ is also one to one.

The semi-direct product on the set $N\times K$ is then the push-back of the group on $G$ via $\psi$. Alternatively, it is the only group law on $N\times K$ making of $\psi$ an isomorphism of groups. Let us write down what it means :

\begin{align} (n_1,k_1)(n_2,k_2)&=\psi^{-1}(\psi((n_1,k_1)(n_2,k_2)))\\ &=\psi^{-1}(n_1k_1n_2k_2)\\ &=\psi^{-1}(n_1 k_1n_2k_1^{-1}k_1k_2)\\ &=\psi^{-1}(n_1\underbrace{k_1n_2k_1^{-1}}_{\in N\text{ because }N\triangleleft G}k_1k_2)\\ &=\psi^{-1}(\underbrace{n_1 k_1n_2k_1^{-1}}_{\in N}\underbrace{k_1k_2}_{\in K})\\ &=(n_1 k_1n_2k_1^{-1},k_1k_2) \end{align}

Whence this gives the definition of the semi-direct product if we fix $\phi:K\rightarrow Aut(N)$ a group morphism then the definition of the semi direct product $N\rtimes_{\phi}K$ is exactly :

$$(n_1,k_1)(n_2,k_2):=(n_1\phi(k_1)(n_2),k_1k_2)$$

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  • $\begingroup$ Thank you for this detail! (Even on my ill-founded question). Seeing the semidirect product as a pushback is really nice. I'm now wondering about what kinds of ways there are in general to make a group out of two subgroups. So the direct product is the most basic, the semidirect product is the pushback of $(n, k) \mapsto nk$ ... are there any more? (I'm also curious about why we need $N$ normal to have $\psi$ be a bijection). $\endgroup$ – Eli Rose Apr 5 '16 at 14:10
  • $\begingroup$ @EliRose, in general, the semi-direct product appears as a very particular case of the decomposition of a group as an extension. We say that a group $G$ is an extension of $Q$ by $N$ if $N$ is normal in $G$ and $G/N$ is isomorphic to $Q$. Which is usually written with a short exact sequence $1\rightarrow N\rightarrow G\rightarrow Q\rightarrow 1$. This extension leads to a semi-direct product if and only if the projection $\pi:G\rightarrow Q$ admits a "section", i.e. a group morphism $s:Q\rightarrow G$ verifying $\pi\circ s=Id_Q$. $\endgroup$ – Clément Guérin Apr 5 '16 at 15:05
  • $\begingroup$ @EliRose, the classification of those extensions uses the second cohomology group of $G$ with values in $Z(N)$ (with an action given by an outer action). If the general case is usually skipped because it involves many complicated calculations, one usually sees the classification of extensions of $Q$ by $N$ when $N$ is abelian (which is more natural than the general case). $\endgroup$ – Clément Guérin Apr 5 '16 at 15:08
  • $\begingroup$ @EliRose, one does not need the $N$ to be normal to have that $\psi$ is a bijection. One only needs it in order to have an interesting group law arising on $N\times K$ (my mistake). $\endgroup$ – Clément Guérin Apr 5 '16 at 15:11

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