Is Group Decomposition Unique in Semidirect Products?

[daveyp225]

Some confusion here reviewing some algebra.

From what I remember, a homomorphism must be specified in order to claim that a group is the semidirect product of H and K, say G \cong H \rtimes_{\phi}\,\, K, where H is normal and \phi:K \to Aut(H) is a homomorphism.

However, I've seen solutions to problems where the subscript is omitted, such as when determining the number of groups of some order. Is it implied then? I always assume it is an inner automorphism, but I know in general that is not true. Which leads me to my next question...

Wikipedia states

"Note that, as opposed to the case with the direct product, a semidirect product of two groups is not, in general, unique; if G and G′ are two groups which both contain isomorphic copies of N as a normal subgroup and H as a subgroup, and both are a semidirect product of N and H, then it does not follow that G and G′ are isomorphic. This remark leads to an extension problem, of describing the possibilities."

But if I have H and H' are isomorphic (via isomorphism f), normal N and N' are isomorphic (via isomorphism g) with G=HN, G'=H'N'. How is G not isomorphic to G'? I would think sending x = hn -> f(h)g(n) = h'n' = x' would be an isomorphism.

 

[micromass]

Some confusion here reviewing some algebra.

From what I remember, a homomorphism must be specified in order to claim that a group is the semidirect product of H and K, say G \cong H \rtimes_{\phi}\,\, K, where H is normal and \phi:K \to Aut(H) is a homomorphism.

However, I've seen solutions to problems where the subscript is omitted, such as when determining the number of groups of some order. Is it implied then? I always assume it is an inner automorphism, but I know in general that is not true. Which leads me to my next question...

Formally, we should never omit the subscript in a semidirect product. The automorphism that we use is very important and should not be omitted. However, there are cases in which the semidirect product is uniquely defined. For example, if I write \mathbb{Z}_3\rtimes \mathbb{Z}_2, then this group is uniquely defined as D_6. All (nontrivial) automorphisms will give rise to this same group.

Wikipedia states

But if I have H and H' are isomorphic (via isomorphism f), normal N and N' are isomorphic (via isomorphism g) with G=HN, G'=H'N'. How is G not isomorphic to G'? I would think sending x = hn -> f(h)g(n) = h'n' = x' would be an isomorphism.

It is not an isomorphism. I'll let you discover why by providing a counterexample. Consider

G=\mathbb{Z}_2\times \mathbb{Z}_3

with H={(0,0),(1,0)} and N={(0,0),(0,1),(0,2)}

and G^\prime=D_6=\{1,a,a^2,b,ab,a^2b\}

with H={1,b} and N=\{1,a,a^2\}. These groups are not isomorphic, but H and H' are and N and N' are. Do you see where your argument fails?

 

[daveyp225]

Formally, we should never omit the subscript in a semidirect product. The automorphism that we use is very important and should not be omitted. However, there are cases in which the semidirect product is uniquely defined. For example, if I write \mathbb{Z}_3\rtimes \mathbb{Z}_2, then this group is uniquely defined as D_6. All (nontrivial) automorphisms will give rise to this same group.
It is not an isomorphism. I'll let you discover why by providing a counterexample. Consider

G=\mathbb{Z}_2\times \mathbb{Z}_3

with H={(0,0),(1,0)} and N={(0,0),(0,1),(0,2)}

and G^\prime=D_6=\{1,a,a^2,b,ab,a^2b\}

with H={1,b} and N=\{1,a,a^2\}. These groups are not isomorphic, but H and H' are and N and N' are. Do you see where your argument fails?

Yes, I see. In general the map won't preserve commutativity. And I assume that even if both were commutative it still won't hold in general? For if we decompose, say, Z/12Z = H x N = Z/4Z x Z/3Z, and G = H'xN' = Z/6Z x Z/2Z? And this is different from Z/6Z \rtimesZ/2Z which would be D12, right? What is G considered then? What about Z/3Z \rtimes Z/4Z?

More questions, haha. If I were to guess, if our group is abelian, by the FTFAG, it has as many different semidirect products as can be retrieved by counting all the cyclic decompositions. Also, simple groups cannot be factored this way per the definition, correct? Besides these, is there a general way to tell how many different semidirect products exist? I would guess by counting the number of distinct automorphisms, but that seems... "hard"

 

[micromass]

Yes, I see. In general the map won't preserve commutativity. And I assume that even if both were commutative it still won't hold in general? For if we decompose, say, Z/12Z = H x N = Z/4Z x Z/3Z, and G = H'xN' = Z/6Z x Z/2Z?

But H and H' aren't isomorphic here? OR do I miss something?

And this is different from Z/6Z \rtimesZ/2Z which would be D12, right? What is G considered then? What about Z/3Z \rtimes Z/4Z?


More questions, haha. If I were to guess, if our group is abelian, by the FTFAG, it has as many different semidirect products as can be retrieved by counting all the cyclic decompositions.

Could you expand here? I have troubles to see what you mean.

Also, simple groups cannot be factored this way per the definition, correct?

Correct.

Besides these, is there a general way to tell how many different semidirect products exist? I would guess by counting the number of distinct automorphisms, but that seems... "hard"

Well, the automorphism groups are well-known and are well studied. So figuring out what the number of automorphisms is isn't really that difficult.
The problem is that some distinct automorphisms give rise to the same semidirect product. That is, two automorphisms \varphi,\psi might be different, but the semidirect products associated with these automorphisms might be equal.

If you want to know the number of different semidirect products then you'll have better luck by looking at the orders of the groups. For example, the semidirect product of \mathbb{Z}_5 and \mathbb{Z}_2 is a group of order 10. But there are only two groups of order 10, so there can only be (at most) two different semidirect products!

 

[daveyp225]

But H and H' aren't isomorphic here? OR do I miss something?
Could you expand here? I have troubles to see what you mean.

I got a little ahead of myself here. It took me some to see the "obvious" fact that if a semidirect product is abelian then it is only a direct product, which happen if the only automorphism is the trivial one. Something I wish was written down somewhere. Though I wonder if all groups which can be decomposed do so as either direct/semi-direct products?

Well, the automorphism groups are well-known and are well studied. So figuring out what the number of automorphisms is isn't really that difficult.
The problem is that some distinct automorphisms give rise to the same semidirect product. That is, two automorphisms \varphi,\psi might be different, but the semidirect products associated with these automorphisms might be equal.

If you want to know the number of different semidirect products then you'll have better luck by looking at the orders of the groups. For example, the semidirect product of \mathbb{Z}_5 and \mathbb{Z}_2 is a group of order 10. But there are only two groups of order 10, so there can only be (at most) two different semidirect products!

Two different semidirect products which includes the trivial (direct) product, right? I have been going though groups of order pq, p<q, today and reading the solutions made me nuts with all the presentation by generators and such. Is it really needed? Is there a simpler way to tell when Z_q \rtimes_f Z_p \cong Z_q \rtimes_g Z_p (this is when we can drop the subscripts I see)? I know there are either 1 or p conjugacy classes through f:Zp->Aut(Zq) iff the order of f(1) divides p and q-1. In the first case we end up with Zpq.

 

[micromass]

I got a little ahead of myself here. It took me some to see the "obvious" fact that if a semidirect product is abelian then it is only a direct product, which happen if the only automorphism is the trivial one. Something I wish was written down somewhere. Though I wonder if all groups which can be decomposed do so as either direct/semi-direct products?

It depends on what you mean with "decomposed". But there are other types of products, like smash products and stuff. Semidirect products aren't the only way of composing groups.

Two different semidirect products including the trivial (direct) product, right?

Right.

I have been going though groups of order pq, p<q, today and reading the solutions made me nuts with all the presentation by generators and such. Is it really needed? Is there a simpler way to tell when Z_q \rtimes_f Z_p \cong Z_q \rtimes_g Z_p (this is when we can drop the subscripts I see)? I know there are either 1 or p conjugacy classes through f:Zp->Aut(Zq) iff the order of f(1) divides p and q-1. In the first case we end up with Zpq.

Well, there are only two groups of order pq. So either we have \mathbb{Z}_{pq} or we have \mathbb{Z}_p\rtimes \mathbb{Z}_q. So if f and g are not the identity, then they induce the same semi-direct product.