1

This is a sequel to $G$ is a group with a normal subgroup $K$ such that $G/K$ is soluble, and $H$ is a nonabelian simple subgroup of $G$, then $H \leq K$ and I'm trying to understand this answer.

The context is that $G$ is a group with a normal subgroup $K$ such that $G/K$ is solvable, and $H$ is a nonabelian simple subgroup of $G$. I believe a canonical projection is any homomorphism $\pi: G \to G/K$.

Now the answerer remarks that:

The image of $H$ in the quotient $G/K$ is $HK/K \cong H/(H \cap K)$

  1. It's not clear to me what $HK$ means. Is is a semi-direct product of $H$ and $K$ (considering $K$ is a normal subgroup and $H$ is not)?

  2. Why is $\pi(H) = HK/K$?

  3. Why is $HK/K$ isomorphic to $H/(H\cap K)$?

1 Answers1

4

  1. $HK$ is defined as $\{hk\mid h\in H, k\in K\}$. Since $K$ is normal, this is a subgroup in $G$, and in particular, it is the smallest subgroup that contains both $H$ and $K$. (In the abelian case, and in particular with vector spaces, you may have seen the notation $U+V$ when $U$ and $V$ are subspaces of some bigger space. That's essentially the same thing.) It is a semidirect product in case $|H\cap K| = 1$.

  2. $HK/K$ is, by definition, the set $$\{hkK\mid h\in h, k\in K\} = \{hK\mid h\in H\}\subseteq G/K$$It is quite clear that the right side description is equal to $\pi(H)$, as $\pi(h) = hK$.

  3. There is a homomorphism $H\to HK/K$ given by $h\to hK$. We can show that it is surjective, and that its kernel is $H\cap K$, which induces an isomorphism $H/(H\cap K)\to HK/K$.

Arthur
  • 199,419
  • How are we sure that $HK$ is a group though? You just defined the set-theoretic product of $H$ and $K$. Wikipedia says The product of two subgroups S and T of a group G is itself a subgroup of G if and only if ST = TS. So we'd have to verify $HK = KH$, but I guess that's true because $K$ is normal? –  May 11 '20 at 15:41
  • 1
    @Triskelion This indeed follows from the normality of $K$. Here is one approach: Take an arbitrary product $h_1k_1h_2k_2$ of two elements in $HK$. We have that $k_1h_2$ may be written as $h_2k_1'$ for some $k_1'\in K$ by normality of $K$ (which says that $h_2K = Kh_2$), so the product is equal to $(h_1h_2)(k_1'k_2)\in HK$. The existence of inverses is proven similarily, as $(hk)^{-1} = k^{-1}h^{-1}$. – Arthur May 11 '20 at 15:44
  • 1
    @Triskelion But your approach also works. We have $$HK = \bigcup_{h\in H}hK = \bigcup_{h\in K}Kh = KH$$where the middle equality follows from normality of $K$. – Arthur May 11 '20 at 15:47
  • Ah, I see! And could you give me a quick argument to see that the map $h \mapsto hK$ is necessarily surjective? I guess for any $hK$ we can pick a corresponding $h$? (But my argument doesn't really seem rigorous. Maybe I'm missing something.) –  May 11 '20 at 15:49
  • 1
    @Triskelion First we need the note from point 2. that $hkK = hK$. And then yes, for each coset $hK\in HK/K$, pick a representative $h$ (which we implicitly have to do to even write down a coset, but that's a notational thing), and that's it. – Arthur May 11 '20 at 15:50
  • This is awesome. Thank you so much! :) –  May 11 '20 at 15:50
  • @Triskelion In general. $HK$ is a subgroup, for subgroups $H$ and $K$, if and only if $HK=KH$ as sets (fairly common beginner exercise; if you have never done it you should try it); this holds, in particular, whenever one of the subgroups is normal, since $hK=Kh$ is true for each $h\in H$; but it may hold even if neither is normal, and even if neither normalizes the other. – Arturo Magidin May 12 '20 at 00:35