# Thread: Simple 1st isomorphism thm question.

1. ## Simple 1st isomorphism thm question.

Sanity check.

We have a surjective homomorphism φ: G → H. We know Ker φ ≅ ℤ and H ≅ ℤ. Then it follows G ≅ ℤ ⊕ ℤ, right?

2. ## Re: Simple 1st isomorphism thm question.

Originally Posted by subfallen
Sanity check.

We have a surjective homomorphism φ: G → H. We know Ker φ ≅ ℤ and H ≅ ℤ. Then it follows G ≅ ℤ ⊕ ℤ, right?
No, that does not follow. It is true that G can be identified with the set $\mathbb{Z}\times\mathbb{Z}$, but the group operation can be different. In $\mathbb{Z}\times\mathbb{Z}$, the usual group operation is given by $(m,n)\oplus(p,q) = (m+p,n+q).$ But you can also define a group operation by $(m,n)\boxplus(p,q) = (m+(-1)^np,n+q).$ The homomorphism from $\mathbb{Z}\times\mathbb{Z}$ to $\mathbb{Z}$ is given in both cases by $\varphi(m,n) = n.$

In general, if $\phi:G\to H$ is a surjective homomorphism with kernel K, then G is (isomorphic to) a semidirect product of K and H. The above example is the semidirect product $\mathbb{Z}\rtimes_\alpha\mathbb{Z}$, where the action $\alpha$ of $\mathbb{Z}$ on itself comes from the automorphism $n\mapsto -n$ of the additive group $\mathbb{Z}.$

3. ## Re: Simple 1st isomorphism thm question.

Originally Posted by Opalg
In general, if $\phi:G\to H$ is a surjective homomorphism with kernel K, then G is (isomorphic to) a semidirect product of K and H. The above example is the semidirect product $\mathbb{Z}\rtimes_\alpha\mathbb{Z}$, where the action $\alpha$ of $\mathbb{Z}$ on itself comes from the automorphism $n\mapsto -n$ of the additive group $\mathbb{Z}.$
That's not right - you mean G is an extension of K by H (or H by K, I can never remember). It is an semidirect product if this extension splits.

An example of a group which is an extension of two groups H and K but is not a semidirect product of H with K is the group,

$G=\langle x, y, z; xy=yx, y^2=1, z^2=1, zxz=x^{-1}y, yz=zy\rangle$.

It contains a (normal) subgroup isomorphic to $\mathbb{Z}$ of finite index, so it is an extension of $\mathbb{Z}$ by a finite group. But it does not split. See here for an explanation as to why.

4. ## Re: Simple 1st isomorphism thm question.

Thanks Swlabr. (You can tell that I am an analyst, not an algebraist.)

But of course it is the semidirect product construction that is needed to answer the question in this thread.

5. ## Re: Simple 1st isomorphism thm question.

Thanks guys. I have a follow-up then. My question came from a Mayer-Vietoris exact sequence including:

... 0 → ℤ → G →φ→ ℤ⊕0 → 0 ...

Where G is not known, and φ: x ↦ (x,-x). So can I conclude anything from this?

6. ## Re: Simple 1st isomorphism thm question.

Ok, I got this response from another board and thought I'd copy it here:

In general if you have a surjective morphism G->H with kernel N, then G is a semidirect product of N and H, and it is not necessarily the direct product.

In your case where N and H are ℤ, the 2 elements of Aut(ℤ) (Id and -Id) give 2 possibilities : the first one is ℤ ⊕ ℤ, and the second one is a group generated by two elements a and b of infinite order with the relation b a b^-1 = a^-1.

However if you already know that G is commutative then it must be the direct product.

7. ## Re: Simple 1st isomorphism thm question.

Originally Posted by subfallen
Ok, I got this response from another board and thought I'd copy it here:
Again, this answer is not accurate! The first and last lines are wrong. Take, for example, $\mathbb{Z}$ under addition. Let $N=\langle n\rangle$ and let $H=C_n$, the cyclic group of order $n$.

It is an easy proof to show that $N$ is normal, and is isomorphic to $\mathbb{Z}$, and that if we quotient out by $N$ we get $H=C_n$. Thus, according to this answer, you have that $\mathbb{Z}$ is a semidirect product of $\mathbb{Z}$ and $C_n$. Therefore, $\mathbb{Z}$ contains a copy of $C_n$ (indeed, it must be $\mathbb{Z}\times C_n$ by the last line!). This is clearly false, as $\mathbb{Z}$ is torsion free (every element is a power of the generating element).

If you have two groups $N$ and $H$ and $G$ is another group such that $G/N\cong H$ then all that can be said is that $G$ is an extension of $H$ by $N$. For an extension to be a semidirect product you need it to split.

8. ## Re: Simple 1st isomorphism thm question.

Originally Posted by Opalg
Thanks Swlabr. (You can tell that I am an analyst, not an algebraist.)

But of course it is the semidirect product construction that is needed to answer the question in this thread.
I have been having a think...is it obvious that the only examples are semidirect products?