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Question: $G=S_5$ and $H=A_4$ and $K=⟨ (12345)⟩$ then show that $HK=A_5$. Where, $HK=\{hk:h\in H,k\in K\}$.

My attempt: I know that, $H$ and $K$ are subgroups of $G$.

First, I will need to show that $HK$ is a subgroup of $G$. (I know the general result that, If $G$ is a group and $H$, $K$ are subgroups of $G$ then $HK$ is subgroup of $G$ if and only if $HK=KH$)

But, I am not able to show that $HK=KH$ here

After this we just need to determine the order of $HK$, which is nothing but equal to $\frac{o(H) o(K)}{o(H\cap K)}$. Clearly it is $60$ (since $H\cap K=\{e\}$)

Now as $HK$ is a subgroup of order $60$ of $G=S_5$, hence we must have $HK=A_5$

So can you help me in the part, where we need to show that $HK$ is a subgroup of $G=S_5$. (Doing it manually i.e. verifying $HK$ is closed under the operation of $G$, takes too much time)

General Mathematics
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1 Answers1

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$H$ is a subset of $A_5$.

$K$ is a subset of $A_5$.

Therefore, $HK$ is a subset of $A_5$, since $A_5$ is a subgroup.

So if you already know that $HK$ contains $60$ elements, then.. ?

testaccount
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  • Then What.... Can you verify the closure property for that 60 elements? Isn't it takes atleast 2-3 days. Can you do it? – General Mathematics May 31 '24 at 15:07
  • Take $S_3$. $H={(1), (12)}$ and $K={(1), (23)}$ both are subgroups of $S_3$ and $HK={(1), (12), (23), (132)}$ is, a subset of $S_3$ and it has four elements! But it is Not a subgroup of $S_3$ – General Mathematics May 31 '24 at 15:10
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    Yes, but in your example $HK$ contains $4$ elements, not $|S_3| = 6$ elements. Here $A_5$ contains $60$ elements, so $A_5 = HK$. You don't need to verify the closure property of anything in this case. – testaccount May 31 '24 at 15:19
  • But, still I can't understand! How can we directly assume that $HK$ is a subgroup of $G$? We don't even need to prove it? Like we need verify that $HK=KH$ or we need to verify that $HK$ is closed under the operation of $G$. – General Mathematics May 31 '24 at 15:28
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    @GeneralMathematics How many subsets of 60 elements has $A_5$ got? Which of those might $HK$ be? FWIW I upvoted this answer. – Jyrki Lahtonen May 31 '24 at 15:28
  • @JyrkiLahtonen Sir, I got you point. But How do we know that set $HK$ contains 60 distinct elements? Isn't the formula that I used in the post, is for order of subgroup $HK$ ? If we use that formula, Isn't that means, we already assumed that $HK$ is a subgroup of $G$? Or we need to compute those elements by hand? – General Mathematics May 31 '24 at 15:36
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    No. That formula is for the number of distinct products $hk$ with $h\in H, k\in K$. It applies for example also to the case of two subsets of two elements of $S_3$ you used as an example above. The proof is simple: $hk=h'k'$ if and only if $h'^{-1}h=k'k^{-1}$. Both sides of the latter equation must be elements of $H\cap K$. Here $H\cap K$ is always a subgroup, so Lagrange shows that in this case $H\cap K={1}$. – Jyrki Lahtonen May 31 '24 at 15:39
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    Indeed as Jyrki points out the formula $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$ just needs $H$ and $K$ to be subgroups of $G$. No assumption that $HK$ is a subgroup is needed. – testaccount Jun 01 '24 at 03:41
  • An alternative approach is with some basic permutation group theory. In $A_5$, the subgroup $H = A_4$ is the stabilizer of a point in ${1, 2, 3, 4, 5}$. Thus for a subgroup $X \leq A_5$, we have $A_5 = HX$ if and only if $X$ is transitive on ${1,2,3,4,5}$. It is clear that $K = \langle (12345) \rangle$ is transitive, so $A_5 = HK$. – testaccount Jun 01 '24 at 03:45