First, I am sorry if this post has been posted here before since I cannot find anything related to it.

I read on wikipedia about the equivalent conditions for semidirect product


The theorem is :

Let $G$ be a group with identity element $e$, a subgroup $H$ and a normal subgroup $N$ (i.e., $N ◁ G$).

With this premise, the following statements are equivalent:

1) $G = NH$ and $N ∩ H = \{e\}$.

2) Every element of $G$ can be written in a unique way as a product $nh$, with $n \in N$ and $h \in H$.

3) Every element of $G$ can be written in a unique way as a product $hn$, with $h \in H$ and $n \in N$.

4)The natural embedding $H → G$, composed with the natural projection $G → G / N$, yields an isomorphism between $H$ and the quotient group $G / N$.

5)There exists a homomorphism $G → H$ that is the identity on $H$ and whose kernel is $N$.

If one (and therefore all) of these statements hold, we say that $G$ is a semidirect product of $N$ and $H$

Could any one give me hints for the directions or what is the best way to prove the equivalents ?

I got the easy one , which 1 to 2.

Thank you all


Let's show $(5) \Rightarrow (1)$. Let $g \in G$. By assumption we have a homomorphism $\phi:G \rightarrow H$ that is the identity on $H$. If $\phi(g)=1$ then $g \in N$. Otherwise $\phi(g)=h$ for some $h \in H$, so $h^{-1}g \in Nh$, hence $g \in Nh$. So $G=NH$. To show $N \cap H =\{e\}$, suppose $g \in N \cap H$. As $\phi$ acts as the identity on $H$ and $g \in N$, we must have $g=\phi(g)=e$.

$(3) \Rightarrow (5)$ holds, as the map $\phi:G \rightarrow H$ defined by $\phi(hn)=h$ is a homomorphism satisfying the required properties (check!).

$(2) \Rightarrow (3)$ holds as $NH=HN$ ($N$ is normal).

$(1) \Rightarrow (4)$ is clear as $G=HN$ and $Nh=Nh' \Leftrightarrow h=h'$ as $N \cap H= \{e\}$

Finally, we show $(4) \Rightarrow (2)$. $(4)$ says that $G/N$ is $\{Nh|h \in H\}$ with $Nh=Nh' \Leftrightarrow h=h'$. So $G=NH$ and for $g \in G$, $g=nh=n'h' \Leftrightarrow h=h'$. So $n=n'$, hence such an expression is unique. $\square$

  • $\begingroup$ this is nice way to show the equivalent. thank you $\endgroup$ – Leonardo Mar 7 '15 at 19:57

Difficult to say what's best, but let's try something.

(2) and (3) are easily seen to be equivalent, by considering inversion.

(1) is easily seen to be equivalent to (2).

(1) implies (5) implies (4) should not be difficult.

To close the circle, I have a not particularly entertaining argument - there's surely something better.

Now assuming (4), note that each element $g N$ of $G/N$ can be written as $g N = h N$ for a unique $h \in H$. So each $g \in G$ can be written as $h n$, for a unique $h \in H$, and some $n \in N$. If we have $h_{1} n_{1} = h_{2} n_{2}$, with $h_{i} \in H$, and $n_{i} \in N$, with $h_{1} \ne h_{2}$, then $1 \ne h_{1}^{-1} h_{2} = n_{1} n_{2}^{-1} \in N$, which contradicts the fact that $H \to G \to G/N$ is a bijection. So (2) holds.

  • $\begingroup$ Really easy proof from (4) to (2). I did not pay attention for every element in $gN$ has the form $hN$. $\endgroup$ – Leonardo Mar 7 '15 at 20:02

This is not complete, but here are some more ideas. For $(2)\implies (1)$ should be easy as well, and showing $(1)\iff (3)$ would be similar. For $(1)\implies (4)$ note that $H\cap N$ implies, that $h_1N=h_2N\iff h_1=h_2$. For $(2)\implies (5)$ use the map $nh\mapsto h$. This gives $(1)\implies (2)\implies (5)$.


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