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My attempt : Suppose $H$ is another subgroup of $S_5$ of order $60$

by Lagrange's Theorem $\vert H\cap A_5\vert ||H|$, then $|H \cap A_5| \in\{1,2,3,5,6,10,20,30,60\}$

$$\Rightarrow |HA_5| =\frac{|H||A_5|}{|H \cap A_5|}=3600,1800,1200,900,720,600,360,180,120,60$$

So, the only candidates for the order of $|H \cap A_5|$ are $30$ and $60$.

Then I want to claim that the order must be $60$. thus $|H \cap A_5|=|A_5|$ $$\Rightarrow H=A_5$$

But I didn't know how to show the former case doesn't hold. Please give me a hint please!

PS. I haven't learned the concepts of normal group and simple group. So, I would like you to explain without using these concepts! Thank you.

the_fox
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fivestar
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    +1 for your effort, especially getting that far without normality. – Randall May 04 '18 at 11:33
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    See https://math.stackexchange.com/questions/159077/how-to-prove-that-a-5-has-no-subgroup-of-order-30 for ideas – lhf May 04 '18 at 11:43

1 Answers1

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  1. Let $H$ a subgroup of index $2$ in $G$ a finite group. Let $g\in G\setminus H$. The subset $g H$ does not intersect $H$. So it is included in $G\setminus H$. Looking at cardinalities, we conclude $g H= G\setminus H$. Since $g\not \in H$, we have $g\cdot g \not \in gH$. That implies $g\cdot g\in H$. Conclusion: the square of any element of $G$ lies in $H$.

  2. Let $G = S_n$. Let's show that every element of $A_n$ is a product of squares of elements of $S_n$. It's enough to show that the product of two transpositions is a square. If the transpositions $(a,c)$, $(b,d)$ have disjoint supports we check $(a,c)(b,d)=(a,b,c,d)^2$. The other case: $(a,b)(a,c)=(c,b,a)=(c,a,b)^2$.

  3. Let $H$ be a subgroup of index $2$ in $S_n$. From 1. we get: $H$ contains all products of squares. Using also 2. we conclude $H\supset A_n$. Therefore $H=A_n$.

orangeskid
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