$HK$ is a subgroup where $H \subseteq N_G(K)$

Question

Let $H, K$ be subgroups of $G$. Let $HK=\{hk: h \in H, k \in K\}$ and $H \subseteq N_G(K)$.

I'm a bit stuck on something. It makes me a bit nervous that $H \subseteq N_G(K)$. Does this mean that $HK=KH$ or for every $h \in H$ and $k \in K$ there is an $h' \in H$ such that $hk=kh'$. I saw this, but in this question it has that $K$ is a subgroup.

$K$ is also a subgroup here. I don't see the difference between the questions. — Tobias Kildetoft, Nov 12 '13 at 19:56
It is a subgroup of $G$ and therefore also of any subgroup of $G$ it is contained in. — Tobias Kildetoft, Nov 12 '13 at 20:02
Sorry. $H$ is a subgroup of $G$. But, $H$ is also a subset of $N_G(K)$. This is why I'm worried. — emka, Nov 12 '13 at 20:04
Try to prove it then, @jdla, usign that $;AB\le G\iff AB=BA;$ . Normality of one of the factors helps a lot here. — DonAntonio, Nov 12 '13 at 20:05
AS Tobias told you, @jdla, a subgroup of $;G;$ is also a subgroup of any other subgroup that contains it, not merely a subset. — DonAntonio, Nov 12 '13 at 20:06
What I'm worried about is that subset assumption. For completeness I just included the rest of the question I'm working on. — emka, Nov 12 '13 at 20:07
I don't see how that subset assumption makes you worry. What about it makes you worry, given the things already mentioned here? — Tobias Kildetoft, Nov 12 '13 at 20:10
If you continue calling it "subset assumption" then either you didn't understand what's been told to you in the comments, or else you don't quite accept it or else you're not careful enough to write. — DonAntonio, Nov 12 '13 at 20:12
And precisely that assumption is what makes so straightforward that $;HK\le G;$ ...! — DonAntonio, Nov 12 '13 at 20:13

score 0 · Answer 1 · answered Nov 12 '13 at 21:08

$N_G(K)$ is the normalizer in $G$ of the subgroup $K$. It's the set of all $g$ such that $gKg^{-1} = K$.

So since $H \subset N_G(K)$, all of $h \in H$ also commute with $K$ like the $g$ above. So $hK = Kh, \ \forall h \in H$ Take the union over $H$ to show that $HK = KH$.

QED

score 0 · Answer 2 · answered Nov 12 '13 at 21:17

Does not

$N_G(K) = \{g \in G \mid g^{-1} K g = K \}? \tag{1}$

Indeed it does; now for $h_1, h_2 \in H$ and $k_1, k_2 \in K$ we wish to establish that

$h_1k_1h_2k_2 \in HK, \tag{2}$

so we follow (1) and note that since

$h_2 \in H \subseteq N_G(K) \tag{3}$

we have

$h_2^{-1}Kh_2 = K, \tag{4}$

which readily implies

$h_2^{-1}k_1h_2 \in K, \tag{5}$

whence there exists $k_3 \in K$ such that

$h_2^{-1}k_1h_2 = k_3, \tag{6}$

or

$k_1h_2 = h_2k_3, \tag{7}$

whence

$h_1k_1h_2k_2 = h_1h_2k_3k_2 \in HK, \tag{8}$

which shows that $HK$ is multiplicatively closed. If $H$ and $K$ are finite, so that $HK$ is also finite, we are done, since a finite multiplicatively closed subset of any group is a subgroup. In the event that $HK$ is infinite, we need to establish that the identity $e \in G$ is in $HK$ and that $(hk)^{-1} \in HK$ for $h \in H, \, k \in K$. Well, $e \in HK$ is virually self-evident, since $H, \, K$ being subgroups implies $e \in H, \, K$; if $hk \in HK$, then $(hk)^{-1} = k^{-1}h^{-1}$; but again, since

$h^{-1} \in H \subseteq N_G(K), \tag{9}$

we find that

$hk^{-1}h^{-1} = (h^{-1})^{-1}k^{-1}h^{-1} \in K, \tag{10}$

and thus we must have $k_4 \in K$ with

$hk^{-1}h^{-1} = k_4, \tag{11}$

whence

$k^{-1}h^{-1} = h^{-1}k_4 \in HK, \tag{12}$

showing that indeed $(hk)^{-1} \in HK$. This completes the proof that $HK$ is a subgroup of $G$ under the given hypothesis. QED

Hope this helps. Cheerio,

and as always,

Fiat Lux

$HK$ is a subgroup where $H \subseteq N_G(K)$

2 Answers2