Is there any example of two subgroups $H, K \le G$, none of which normal in $G$, such that $G = HK$?

  • $\begingroup$ It is a theorem that $HK$ is a subgroup of $G$ (and $G$ itself is certainly a subgroup) if and only if (at least) one of $H,K$ is normal, so you may have a long search... $\endgroup$ Aug 30, 2015 at 2:58
  • 2
    $\begingroup$ @DavidWheeler I have proved HK is a subgroup of G iff HK=KH. But I don't think it imply in anyone way result you stating. $\endgroup$
    – Sushil
    Aug 30, 2015 at 3:09
  • $\begingroup$ Look at the alternating group $A_5$. $\endgroup$ Aug 30, 2015 at 13:45

2 Answers 2


Yeah, there are examples. Take the 5-cycle $\sigma=(1,2,3,4,5) \in A_5$ and let $H= \langle \sigma \rangle \subseteq A_5$. Naturally $A_4$ can be considered as subgroup of $A_5$ (it's the stabilizer of $5$). Then you have

$$A_5 = H A_4.$$

This can be shown by showing the injectivity of the map $H \times A_4 \to A_5$, $(\sigma^k,\tau) \mapsto \sigma^k\tau$: $\sigma^{k_1}\tau_1=\sigma^{k_2}\tau_2$ implies $\sigma^{k_1-k_2} = \tau_2 \tau_1^{-1} \in H \cap A_4 = \{e\}$). Surjectivity follows then by cardinality (you have $|H|=5$, $|A_4|=12$ and $|A_5|=60$).

But since $A_5$ is simple neither $H$ nor $A_4$ is a normal subgroup of $A_5$.


The following is a systematic way of constructing such examples:

Let $A$ be the ring of (strictly) upper triangular matrices of degree $n \geq 3$, with entries in an arbitrary field. Then, the underlying set $A$ forms a group with respect to the operation $x\circ y=x+y+xy$. Denote this group by $A^{\circ}$ and the additive group of $A$ by $A^+$.

The group $A^{\circ}$ acts on $A^+$ via $x^g=x+xg$, where $x \in A^+$ and $g \in A^{\circ}$. Let $G=A^{\circ} \rtimes A^{+}$.

Now, $G$ is a product of the subgroups $H=\{(x,0)\,|\, x\in A^{\circ}\}$, and $K=\{(x,x)\,|\, x\in A\}$. Indeed, any element $(x,y) \in G$ can be written as: $$(x,y)=(x\circ y^{-1},0)(y,y).$$

I think you can, safely, complete the assertions without proofs.

Remark. with some trivial exceptions, you can replace $A$ by any nilpotent ring, or more generally by a radical ring (i.e., a ring in which the circle operation defined above is a group law).

We have $G=LH$, and $G=HK$; where $L=\{(0,x)\,|\, x\in A\}$. Still there is a third factorization!!


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