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I'm proving that if $H$ is a subgroup of a group $G$ and $K$ is a normal subgroup of $G$, then $HK$ is a subgroup of $G$.

I've tried pondering the fact that $HK$ is a subgroup of $G$ iff $HK = \langle H \cup K \rangle$ but I'm not sure how I'd use that.

I know that $H \cup K$ is a subgroup of $G$ because $H$ and $K$ are (EDIT: This is false)...Other than that, I'm not sure how to proceed.

user3200098
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    Note that it is generally not the case that the union of two subgroups is itself a subgroup. – amWhy Mar 04 '14 at 21:56
  • The only thing that you need to check is that $HK$ is closed under the group operation. How would you go about doing that? – Joshua Pepper Mar 04 '14 at 21:56
  • The union of two subgroups $H$ and $K$ is a subgroup if and only if either $H\leq K$ or $K\leq H$. (For a reference, see this question) – user1729 Mar 04 '14 at 21:58
  • Hm, was it the intersection that was always a subgroup? – user3200098 Mar 04 '14 at 22:02
  • It is true that the intersection of two subgroups is always a subgroup (note that it might be trivial). – Joshua Pepper Mar 04 '14 at 22:06
  • Yes, the intersection is always a subgroup. (The issue with unions is that they will not be closed under multiplication. For example, take two subgroups of the Klein 4-group each of which have order two, and are non-equal. They union to give you a set with three elements, which therefore cannot be a subgroup. The issue is that the two non-trivial elements of these two subgroups multiply to give you the fourth element of the Klein 4-group) – user1729 Mar 04 '14 at 22:06

3 Answers3

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Hint: Use normality to give you $kh=hk_1h^{-1}h$.

user1729
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  • I understand you correctly?

    $H$ is normal subgroup of $G$ so $H = aHa^{-1}$ which implies that $aha^{-1} \in H$ which further implies $khk^{-1} \in H$...So naturally $hkhk^{-1} \in H$...?

    I'm not quite understanding the subscript on one of the $k$'s though

     – user3200098 Mar 04 '14 at 22:00
  • In your question it is the subgroup $K$ which is normal. So, as $K$ is normal we can conjugate $k$ by $h$ to get an element $k_1$ of $K$, $h^{-1}kh=k_1\in K$. – user1729 Mar 04 '14 at 22:04
  • I'm sorry, I don't understand what you mean? Am I far off with my assumption of your answer? My foundation in this course is horrible, so try to go as easy as you can on the notation, please. – user3200098 Mar 04 '14 at 22:06
  • The first thing you have to realise is that in my answer (and your original post) it is the subgroup $K$ which is normal, not the subgroup $H$. The second thing (the main thing) is that because $K$ is normal we have that $h^{-1}kh\in K$ for our specific $K$, and all this means is that there exists some $k_1\in K$ such that $h^{-1}kh=k_1$. Does that make sense? – user1729 Mar 04 '14 at 22:09
  • Yes this does make sense. I agree, there exists a $hkh^{-1} = k_1 \in K$. (Isn't it $hkh^{-1}$ and not $h^{-1}kh$? The definition reads: $aHa^{-1}$ in my textbook.) – user3200098 Mar 04 '14 at 22:11
  • Also, you are defining $kh$, but I would like to prove that $HK$ (i.e. $hk$) is a subgroup of $G$. Why is this? – user3200098 Mar 04 '14 at 22:16
  • The placing of the inverse doesn't matter, as $h=(h^{-1})^{-1}$. So you can just write $h^{-1}k(h^{-1})^{-1}=k_1\in K$ (and $h^{-1}\in G$). Anyway, the next thing you need to realise is that because of this you can write $k=h^{-1}k_1h$. This is the working which gives you my hint. – user1729 Mar 04 '14 at 22:17
  • The purpose of my hint is to prove that $HK$ is closed under multiplication and inverses. This proves the result. I will leave it to you now to join up the dots - I need my bed! – user1729 Mar 04 '14 at 22:18
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Use $hK=Kh$ to obtain $HK=KH$.

Pedro
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Let $h,j \in H$. Given $K$ is normal, $k\in K \implies xkx^{-1}\in K$ for all $x\in G\implies xK=Kx$.

  1. Closure. $hK, jK\subseteq HK\implies hKjK = hjKK = hjK\subseteq HK$.

  2. Identity. $e\in H$ and $e\in K\implies ee = e\in HK$.

  3. Associativity. $HK\subseteq G$ and $G$ is associative.

  4. Invertibility. $(hK)^{-1} = K^{-1}h^{-1}= Kh^{-1} = h^{-1}K \subseteq HK.$

Hence $HK$ is a subgroup of $G$. $\Box$

hchar
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