Show that $A_5$ does not have a subgroup of order 15, 20, nor 30.
I looked up other posts such as the one below, but they used some things we don't cover in the course.
$A_5$ has no subgroup of order 15 and 20
Instead, we did a similar example in class:
Show that $A_4$ has no subgroups of order 6.
Suppose $H \preceq A_4$
with $|H| = 6$. Then $[A_4: H] = 2$ and so $H \unlhd A_4$. Therefore,
$(aH)^2 = H \forall a \in A_4$. This means $a^2 \in H \forall a \in
A_4$. We can compute to show that $A_4$ has 9 elements of the form
$a^2$, and by above they must all be in $H$. But this is a
contradiction since $|H| = 6$. Hence no such $H$ exist.
But the problem with trying to apply a similar proof to my question is that, since $[A_5: H]$ = 4, we cannot say that $H \unlhd A_5$. Hence, we cannot say $(aH)^4 = H$ since we cannot form the factor group $A_5/H$. If that is the case, how else can I solve the question without using simple group and Sylow's Theorem.
Best Answer
So, we've pretty much narrowed it down to the subgroup of order $20$.
To tackle that, note that if $H\le A_5$ with $\mid H\mid=20$, then let $A_5$ act on the set of $3$ left cosets $A_5/H$. We get a homomorphism from $A_5$ into $S_3$, which is non-trivial: it's kernel $\cap_{x\in G}xHx^{-1}\subset H$.
Briefly, the homomorphism is defined by letting $\varphi (g)$ be the element of $S_3$ corresponding to the permutation of the cosets brought about by applying $g$. That's let $\bar{g_1},\bar{g_2},\bar{g_3}$ be the cosets. Multiplying by $g$ on the left permutes them. We get $\bar{g_1}\to \bar {gg_1},\,\bar {g_2}\to\bar{gg_2}$ and $\bar {g_3}\to\bar{gg_3}$. It's easy to see that $g$ takes different cosets to different cosets. So we really do have an element of $S_3$.
Now to continue note that $A_5=\langle(12)(34),(135)\rangle$. And $(12)(34)(135)$ has order $5$. Then $e=\varphi ((12)(34)(135))=\varphi ((12)(34))\varphi ((135))\implies \varphi ((12)(34))=\varphi ((135))=e$, because the orders divide $2$ and $3$ (and are the same). So we've reached a contradiction.