# Understanding a proof by Gallian that $A_5$ has no subgroup of order $30$.

## Details:

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. $$\color{green}{\checkmark}$$

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

## Thoughts:

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

Specifically:

$$\color{red}{X_1}$$: 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.

## 2 Answers

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?