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> Q. Prove that $A\_5$ has no subgroup of order $30.$
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Now at the end of the book he has given an outline of the proof as follows :

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> Suppose that $H$ is a subgroup of $A\_5$ of order $30.$ We claim that $H$ contains all $20$ elements of $A\_5$ that have order $3.$ To verify this, assume that there is some $\alpha$ in $A\_5$ of order $3$ that is not in $H$. Then $A\_5=H \cup \alpha H$. It follows that $\alpha^2 H=H$ or $\alpha^ 2H=\alpha H$. Since the latter implies that $\alpha \in H$, we have that $\alpha^2 H =H$, which implies that $\alpha^2 \in H$. But then $\langle \alpha \rangle= \langle \alpha^2 \rangle \subseteq H.$ Which is a contradiction of our assumption that $\alpha$ is not in $H$. Since $|H|=30$, we have a contradiction.
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In this outline of proof he assumes $\alpha \notin H$ and by following deductions he achieves $\alpha \in H$. This then supposedly implies that our assumption that $\alpha \notin H$ was false and thus every $\alpha$ of order $3$ is in $H$. Isn't it that he should have achieved contradiction in some other fact regarding $A\_5$?

I mean when we prove $A \implies B$ by contradiction, we assume $\sim B$ to be true and after some deductions we get contradiction that $A$ is false rather than $B$ is true. Here we have "$H$ is a subgroup of order $30 \implies$ all elements of $A\_5$ of order $3$ belong to $H$." So we assume there exists such an element not in $H$. In both of the cases that follow, we find that that element is in $H$. I think this is paradoxical somehow. That is $\alpha \notin H \implies \alpha \in H.$

Please help me to get clear picture of what is going on here.