I'm (still) reading "Contemporary Abstract Algebra," by Gallian. My motivation for doing so is mostly recreation.

This is Exercise 7.51 and its solution ibid.

There's a few details in the proof that I don't understand yet, so I'm typing it out step-by-step as a question and if I still don't get it, I'll post it here.

The Proof:

Exercise: Prove that $A_5$ has no subgroup of order $30$.

Proof: Suppose $H\le A_5$ such that $|H|=30$. $\color{green}{\checkmark}$ We claim that $H$ contains all $20$ elements of $A_5$ of order $3$. $\color{red}{X_1}$

To show this, assume $\exists\alpha\in A_5\setminus H$ with $|\alpha|=3$. $\color{green}{\checkmark}$ Then $A_5=H\cup\alpha H$. $\color{green}{\checkmark}$ It follows that $\alpha^2H= H$ or $\alpha^2H=\alpha H$. $\color{green}{\checkmark}$ Since the latter implies $\alpha\in H$, we have $\alpha^2 H=H$, so $\alpha^2 \in H$. $\color{green}{\checkmark}$

But then $\langle \alpha\rangle=\langle\alpha^2\rangle\subseteq H$. $\color{red}{X_2}$

This contradicts our assumption that $\alpha\notin H$. $\color{green}{\checkmark}$

By the same argument, $H$ must contain all $24$ elements of order $5$. $\color{red}{X_3}$

Since $|H|=30$, we have a contradiction. $\color{green}{\checkmark}$

$\square$ $\color{green}{\checkmark}$


Although I have a rough understanding of the structure of a proof, the parts marked with red $X$s are unclear to me.


$\color{red}{X_1}$: I'm not sure why, other than Lagrange's Theorem, there are $20$ such elements.

$\color{red}{X_2}$: Oh, I think this is because $\alpha^2=\alpha^{-1}$.

$\color{red}{X_3}$: My problem here is similar to $\color{red}{X_1}$ (so . . . Use Lagrange?), although I think I understand a lemma that states that there is a multiple of $\varphi(n)$ elements of order $n$ in any group.

Please help :)

NB: I hope this constitutes just one question, so is not too broad; it's about one proof after all.


I suppose you know that an order of a permutation equals to the $lcm$ of the lengths of its disjoint cycles. Since it is only $A_5$ it is not too hard to see that a permutation has order $3$ if and only if it is a $3$-cycle. And from here it is just combinatorics. How can you create a $3$-cycle? First of all you need to choose $3$ elements from the set $\{1,2,3,4,5\}$ which is $\binom {5}3$ options. And once you chose these elements you need to choose the image of each element. All you have to do here is to choose the image of one of these $3$ elements, and because you want your permutation to be a $3$-cycle that will tell you what are the images of the other two elements. So you have just $2$ options here. Hence the number of $3$ cycles is $\binom {5}3\times 2=20$.

Now you can do something similar to find the number of elements of order $5$. As for $X_2$, you understood it right.


$X_2$ is OK.

$X_1, X_3$: these come from the unique decomposition of a permutation as a product of disjoint cycles. It shows the elements of order $3$ are the $3$-cycles, and those of order $5$ are the $5$-cycles. Do you see why there are $20$ and $24$ of them, respectively?


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