# Quotient of group by a semidirect product of subgroups

1. Feb 18, 2016

### mnb96

Hello,

if we consider a group G and two subgroups H,K such that $HK \cong H \times K$, then it is possible to prove that:

$$G/(H\times K) \cong (G/H)/K$$

Can we generalize the above equation to the case where $HK \cong H \rtimes K$ is the semidirect product of H and K?

Clearly, if HK is a semidirect product, then it might not be normal in G, so my guess is that the best we can do is to calculate the quotient $G / \langle H \rtimes K \rangle^G$ where $\langle H \rtimes K \rangle^G$ denotes the normal closure (or conjugate closure) in G of the semidirect product.
Do you have any hint on how to do this?

2. Feb 18, 2016

### Staff: Mentor

Interesting question, although I doubt that it is a valid way because normalizing a subgroup makes the subgroup a normal subgroup in its normalizer but not in the whole group. And if the normalizer is already the whole group then it had been a normal subgroup in the beginning.

I'm not quite sure what $\langle H \rtimes K \rangle^G$ is, e.g. whether it is $H \rtimes K^G.$
A quick calculation gave me that the $K$-component of $\langle H \rtimes K \rangle^G$ is only affected by the $K$-component of the $g \in G$ used for conjugation and thus of no use. I set the multiplication $$(h,k) \cdot (\bar{h}, \bar{k}) = (h \cdot \phi(k)(\bar{h}),k \cdot \bar{k})$$ with a homomorphism $\phi : K → Aut(H)$. However, I might have made a mistake on my back envelope mixing the subgroups so you should correct me if so (I'm a little rusty). I want to figure out where it leads us, too. Maybe there is another trick that could be applied.

Last edited: Feb 18, 2016
3. Feb 18, 2016

### andrewkirk

What is meant by
$$G/(H\times K) \cong (G/H)/K\textrm{?}$$
Presumably H must be normal in G.
But assuming that, the quotient does not appear to be properly defined, as K is not a subgroup of (G/H).
Is it intended to signify (G/H) / (HK/H) ?
If so, can we be sure that H is normal in HK?

4. Feb 19, 2016

### mnb96

Hi. Thanks for both replies.
It seems that I should have formulated better my premises, so I'll make a second attempt:

In the first example involving the direct product we shall assume that have a group G and two subgroups H,K such that:

1) both H,K are normal in G
2) H∩K={1}

We can now say that the subgroup HK is isomorphic to the direct product H×K, thus:

G/(H×K) ≅ G/(HK) ≅ (G/H)/(HK/H) ≅ (G/H)/K

where I used respectively the 2nd and 1st isomorphism theorems in the last two "identities".

Now, my original question was essentially: Can we generalize the above "equation" if we relax assumption 1) by stating that:

1) H is normal in G
2) H∩K={1}

In this case, note that the group HK should be isomorphic to the semidirect product $H \rtimes K$.
Clearly, HK is not necessarily normal in G, so my guess was that the best we could do was to consider its conjugate closure <(HK)G> (which is normal in G) and calculate:

$G/ \langle(H\rtimes K)^G \rangle \cong G/ \langle(HK)^G\rangle \cong \; ?$

5. Feb 19, 2016

### Staff: Mentor

Proof?

6. Feb 19, 2016

### andrewkirk

I don't understand this.
According to my texts, the second identity is an instance of the third isomorphism theorem (not the second) and the third identity is not an identity because (G/H)/K has no meaning. We can give it a meaning by interpreting it as meaning (G/H)/(HK/H) but in that case the third identity says nothing.

7. Feb 19, 2016

### mnb96

If we have to trust Wikipedia, then "The conjugate closure of any subset S [containing the identity element] of a group G is always a normal subgroup of G". By the way, I proved that statement by myself too. Were you interested in the explicit steps of my proof, or were you just questioning the validity of the statement?

You are actually right. When writing (G/H)/K we are basically saying that K is a normal subgroup of the quotient G/H (whose elements are cosets) which doesn't make much sense, so let's stick to G/(H×K) ≅ (G/H)/(HK/H).

Any hint about the remaining part of my question involving semidirect products?

8. Feb 19, 2016

### Staff: Mentor

I was interested in the proof or simply the definition of conjugate closure since I think it differs from the normalisator which I thought you were referring to. Do you mean $K^G = \{gKg^{-1} | g \in G\}$ or $K^G = \{g \in G | gKg^{-1} \subset K \}$? That was why I'm asking.

9. Feb 19, 2016

### andrewkirk

It seems to me that the natural item to put in place of the question mark is $(G/H)\ /\ (\langle(HK)^G\rangle /H)$

To show that's well-defined we need
1. $H$ to be normal in the conjugate closure; and
2. $\langle(HK)^G\rangle /H$ to be normal in $G/H$
The first of these follows easily because the normality of $H$ in the conjugate closure is inherited from its normality in $G$.
The second follows from the normality of the conjugate closure in $G$.
So it's well-defined.

Also, there is no other obvious item to put in that place.

The only (!) thing we have to do is prove isomorphism.

10. Feb 19, 2016

### andrewkirk

It's the first one.
So the conjugate closure $\langle K^G\rangle$ is the subgroup of $G$ generated by that.

11. Feb 19, 2016

### suremarc

@fresh42, the conjugate closure of a subgroup is always normal in the parent group.
Edit: ninja'd by andrewkirk. Oh well

@OP: It's worth noting that non-isomorphic subgroups of a given group can have equivalent conjugate closures. E.g. when a given normal subgroup is simple, any two subgroups thereof have equivalent conjugate closures.
In this sense, taking the conjugate closure of $H\rtimes K$ fails to preserve the structure of $K$. So any statement about $G/\langle(HK)^G\rangle$ would suffer the same loss of information.

12. Feb 19, 2016

### Staff: Mentor

Yes, I see. My uncertainty came from the usage of "normal closure" in the OP which I falsely related to the normalisator.

13. Feb 20, 2016

### mnb96

Thank you all for you help.
I think andrewkirk basically answered my original question. Apparently we can state that:

$$G/ \langle (H\rtimes K)^G \rangle \cong (G/H)/(\langle (HK)^G \rangle / H)$$

and I also think that the 3rd isomorphism theorem (a.k.a. "freshman theorem") guarantees that the above is indeed an isomorphism.

I also went a little bit further and proved that for the rightmost term of the above isomorphism we have that: $\langle (HK)^G\rangle = \langle K^G\rangle \langle H^G\rangle = \langle K^G\rangle H$, thus: $$\langle (HK)^G\rangle / H \cong \langle K^G\rangle$$
This also means that if one would want to use that ambiguous abuse of notation that I used in my first post (which I'd now discourage), then we could write: $$G/\langle(H\rtimes K)^G\rangle \cong (G/H)/\langle K^G\rangle$$
This is of course a bit dangerous because one has to keep in mind that $\langle K^G\rangle$ is to be interpreted as the quotient defined above.

14. Feb 20, 2016

### suremarc

There's one more thing I was hoping you'd see.

$H$ and $\langle K^G\rangle$ both normal in $G$ (and a fortiori normal in $HK^G$) whose intersection is the trivial group. (Can you prove this?) Hence $H\langle K^G\rangle$ is a direct product, and we can apply the lemma stated in OP.

15. Feb 20, 2016

### Staff: Mentor

It would be interesting to know some examples, eventually matrix groups, where $H\rtimes K \lneqq G, H\times K^G \lneqq G$ and $K \neq K^G \lvertneqq G$ are proper subgroups without filling the gaps by simply adding direct summands.
Likewise for $H \rtimes K = G$ which might be easier. However, $K \neq K^G \lvertneqq G$ seems to be a strong condition to me.

If I am correct, e.g. the Poincaré group is a semidirect product and might be a candidate.

16. Feb 20, 2016

### mnb96

Oh! You are absolutely right. I should have probably waited few minutes more before writing here, since all the elements to infer your observation were basically under my eyes. Thanks for letting me notice this.

This actually makes me wonder whether we could have reached the same result by considering the abelianization of the semidirect product HK in order to obtain the direct product H<K>G...(Edit: A very quick calculation (which I will double-check tomorrow) apparently led me to a positive answer.)

Last edited: Feb 20, 2016
17. Feb 20, 2016

### suremarc

I wouldn't say so. Suppose $H\rtimes K$ were normal in $G$, Then in fact $K^G\leq HK$. What's left is to make sure $K$ isn't normal in $HK$.
In general, we need only find a normal subgroup of $G$ which splits into a nontrivial semidirect product. Examples are plentiful--e.g. $\mathrm{A_4<S_4}$ which decomposes as $V4\rtimes C_3$.

As for the Poincaré group, Minkowski spacetime doesn't have any invariant subspaces under the action of the Lorentz group, I think. So the answer would be in the negative.

18. Feb 20, 2016

### Staff: Mentor

The Poincaré group (as defined in Wiki) is $ℝ^{1,3} \rtimes SO(1,3)$. What I don't know is whether there is a natural embedding in some larger group or whether canonical extensions exist. For its Lie Algebra there is a tower of semidirect extensions. However I have no idea whether similar constructions exist on the group level.

19. Feb 21, 2016

### suremarc

Well, $\mathrm{Lie}(G)=\mathrm{Lie}(G_0)$, so a tower of extensions for $\mathrm{Lie}(G)$ corresponds to a tower of group extensions for $G_0$, right? (Just speculating based off what I've read--not an expert with Lie theory.)

20. Feb 21, 2016

### Staff: Mentor

Not necessarily because a semidirect product always involves an operation, group or algebra. However a representation of a Lie algebra doesn't need to have a correspondence on the group level. At least I don't know a general mechanism for it. Going tangential is somehow easier than the other way. In case of the Poincaré group I'd like to know whether there are more translations extending the group and have a physical meaning, some sort of $ℝ^{1,3} \rtimes (ℝ^{1,3} \rtimes SO(1,3))$. In the OP the essential question was about the differences between direct and semidirect products with respect to the factors. That's why the Poincaré group came to my mind.