Is semidirect product unique?

This is about semi direct product on Dummit and Foote algebra text book. Why is this statement true?

Theorem 12. Suppose $G$ is a group with subgroups $H$ and $K$ such that

1. $H\trianglelefteq G$, and
2. $H\cap K=1$

Let $\varphi:K\to\operatorname{Aut}(H)$ be the homomorphism defined by mappinf $k\in K$ to the automorphism of left conjugation by $k$ on $H$. Then $HK\cong H\rtimes K$. In particular, if $G=HK$ with $H$ and $K$ satisfying $(1)$ and $(2)$, then $G$ is the semidirect product of $H$ and $K$.

I think semidirect product of $H,K$ depends on the choice of homomorphism $\varphi$.

But once $H,K$ is determined, $HK$ is unique, I think.

For example, if $H =\mathbb{Z}/3\mathbb{Z}$ and $K=\mathbb{Z}/4\mathbb{Z}$ then $Aut(H)=\mathbb{Z}/2\mathbb{Z}$.

Then homomorphism from $K$ to $Aut(H)$ is not unique. (Namely, there are two cases.)

So semidirect product of $H,K$ is not unique.

So how can I understand this situation?

Maybe, $HK$ is not unique, and there is a theorem such that $HK$ is isomorphic to the semidirect product of $H,K$ for some homomorphism $\varphi$?

Thanks!

• once $H,K$ are determined (as subgroups of some group $G$) then $HK$ is unique. but knowing $H,K$ as groups only, $HK$ is not always uniquely determined; even it doesn't make sense. If we have given $\varphi\colon K\rightarrow H$ a homomorphism, then there is a way to determine $HK$ uniquely. – p Groups Dec 29 '15 at 8:03
• Note that the notation $HK$ doesn't even make sense if $H$ and $K$ aren't already subgroups of some group $G$. To produce (outer) semidirect products out of two abstract groups $H$ and $K$, you need the data of a homomorphism $K \to \operatorname{Aut}H$. Conversely, each semidirect product of $H$ and $K$ yields a homomorphism $K \to \operatorname{Aut}H$, and any two semidirect products that yield the same homomorphism are isomorphic (they are each isomorphic to the outer semidirect product constructed through the homomorphism). – Alex Provost Dec 29 '15 at 8:06

(A) You are right: semi-direct product of $H$ by $K$ depends on choice of homomorphism $\varphi\colon K\rightarrow Aut(H)$.

(B) When we start with group $G$, and subgroups $H,K$ with

(1) $H\trianglelefteq G$,

(2) $HK=G$,

(3) $H\cap K=1$,

then (the known group) $G$ is called as internal semi-direct product of $H$ by $K$. Here, we already know that $H$ is normal in $G$, hence every element $k\in K$ determines an automorphism of $H$: $h\mapsto khk^{-1}$. This gives a homomorphism $\varphi\colon K\rightarrow Aut(H)$.

(C) Suppose we don't know $G$, but we start with groups $H$ and $K$ (not subgroups), with a homomorphism $\varphi\colon K\rightarrow Aut(H)$ [like, in your question, see after ..for example...]. Then we can construct a group $G$ such that (roughly) $H$ and $K$ satisfy (1), (2) and (3) above. In this case, the constructed group $G$ is called as external semi-direct product of $H$ by $K$.

• Thanks. But I'm still confused. I've been reading about classification of finite group of small order. Here is what is confusing. If We know H,N and suppose HN = G.(Think H,N as a sylow subgroup of G). then yes, G can be Semidirect product of H,N for some corresponding homomorphism from H to Aut(K). i.e. For given (phi), G is determined. But Is there any other possibility that G is not isomorphic to the semidirect product of H and N, but G is isomorphic to HN? – nicksohn Dec 29 '15 at 7:27
• To make clear your last question: let me know, what is given complete data (hypothesis) in your question, and what is expectation? – p Groups Dec 29 '15 at 7:29
• Ah...! (B) solve my problem. For Given H,K, that gives (Phi) and for that (phi) HK isomorphic to the semiproduct of H and K. Thus, HK should be isomorphic to the semiproduct of H and K for some (Phi). So we only consider the case all possible (Phi), and correspoding Semidirect product of H and K. and There are no other options! Thanks! – nicksohn Dec 29 '15 at 8:26

I find it helpful to distinguish between the inner semidirect product and the outer semidirect product. (Similiarly to the distinction between the inner and the outer direct sum of vector (sub)spaces.)

Given a group $G$ and two subgroubs $H, N \subseteq G$, then $G$ is called the inner semidirect product of $H$ and $N$ with $N$ normal, if $NH = G$, $H \cap N = 1$ and $N$ is normal in $G$. We then write $G = N \rtimes H$. (This does not depend on any homomorphism.)

Given any two groups $N$ and $H$ and a group homomorphism $\theta \colon H \to \mathrm{Aut}(N)$ the corresponding outer semidirect product $N \rtimes_\theta H$ is defined as the set $N \times H$ together with the multiplication $$(n_1, h_1) \cdot (n_2, h_2) = (n_1 \theta(h_1)(n_2), h_1 h_2).$$ Notice that this every much depends on $\theta$.

Now what is the connection between these two?

If $G$ is a group and $H, N \subseteq G$ are subgroups such that $G = N \rtimes H$, then $N$ is normal and thus invariant under the conjugation by $H$. So we get a group homomorphism $\theta \colon H \to \mathrm{Aut}(N)$ with $\theta(h)(n) = hnh^{-1}$. It then follows from direct calculations that the map $$N \rtimes_\theta H \to G = N \rtimes H, \quad (n,h) \mapsto n \cdot h$$ is a group isomorphism. So every inner semidirect product can be seen as an outer semidirect product via the conjugation action of $H$ on $N$.

If on the other hand $N$ and $H$ are groups and $\theta \colon H \to \mathrm{Aut}(N)$ a group homomorphism, then we can regard $N$ and $H$ as subgroups $\hat{N}$ and $\hat{H}$ of $N \rtimes_\theta H$ via the obvious inclusions. It then follows that $\hat{N}\hat{H} = \hat{N} \rtimes_\theta \hat{H}$, that $\hat{N} \cap \hat{H} = 1$ and that $\hat{N}$ is normal in $N \rtimes_\theta H$. So $N \rtimes_\theta H$ is the inner semidirect product of $\hat{N}$ and $\hat{H}$ with $\hat{N}$ normal.

PS: So given any two groups $H$ and $N$ there may be multiple outer semidirect products $N \rtimes_\theta H$ because we can choose different homomorphisms $\theta \colon H \to \mathrm{N}$. But if we are given an inner semidirect product $G = N \rtimes H$ the group homomorphism $H \to \mathrm{Aut}(N)$ is implicitely given: It is hidden in the group structure of $G$ and can be retrieved via the conjugation action of $H$ on $N$.

• Thanks for this nice answer! Very helpful! – nicksohn Dec 29 '15 at 8:26

Definition: A semidirect product of two abstract groups $H$ and $K$ is a group $G$ containing (isomorphic copies of) $H$ and $K$ as subgroups, with $H$ normal inside $G$, and such that $G$ factors uniquely as $G = HK$ (the uniqueness is equivalent to asking that $H \cap K = 1$).

Fact: Any semidirect product of $H$ and $K$ induces a homomorphism $K \to \operatorname{Aut}H$ by conjugation (this works since $H$ is normal).

Theorem (existence and uniqueness): Given abstract groups $H$, $K$ and a homomorphism $\phi:K \to \operatorname{Aut}H$, there exists a semidirect product of $H$ and $K$ ("outer semidirect product") $G = H \rtimes_\phi K$ inducing $\phi$. Moreover, any semidirect product of $H$ and $K$ inducing $\phi$ is isomorphic to $H \rtimes_\phi K$.

Note: Two different homomorphisms $K \to \operatorname{Aut}H$ can induce the same semidirect product. For example, let $K = \mathbb{Z}$, let $H$ be any nonabelian group, say with $h \in H \setminus Z(H)$, and consider the conjugation homomorphism $\phi:\mathbb{Z} \to \operatorname{Aut}H$ defined by $\phi_1(h') = hh'h^{-1}$. By our choice of $h$, $\phi$ is a nontrivial homomorphism, but both $\phi$ and the trivial homomorphism $\mathbb{Z} \to \operatorname{Aut}H$ yield the direct product $H \times \mathbb{Z}$.

• Thanks, A.P. Have a nice day. – nicksohn Dec 29 '15 at 9:06