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Why $A_{5}$ has no subgroup of order 20?

Thanks!

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Tell us what you've done so far. Do you have any idea how to solve this? Are you stuck somewhere? –  Olivier Bégassat Jul 11 '12 at 12:20
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Well, I've tried this: Let us assume $H$ is such a group. $A_{5}$ acts on the group of left cosests of $H$ by left multiplication and therefore $A_{5}$ is isomorphic to a subgroup of $S_{3}$ which is not possible. But it seems too simple, and I've used no special prpoerty of $A_{5}$ so I think I've got it wrong somewhere. –  Roy Jul 11 '12 at 12:39
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up vote 9 down vote accepted

First, yes: it is that simple (ah, how we loathe simple things, uh?)

Second, of course you've used a very special property of $\,A_5\,$: it is a simpe group! Otherwise how could you deduce $\,A_5\,$ is isomorphic to a subgroup of $\,S_3\,$?? After all, it could be that the kernel of the action you described is big enough as to make this possible, but that kernel is a normal subgroup of $\,A_5\,$, so...

This exercise is more or less middle mathematics undergraduate level (unless, of course, somebody says it is high school's level...we've had some wise guys like these around lately), so you must give more credit to yourself when you reach logically some result. Just be sure every single claim you make you can back it.

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Thank you DonAntonio. But now I see that I cannot back the claim that $A_{5}$ is isomorphic to a subgroup of $S_{3}$. I thought that the following is true: If a group $G$ acts on the set of left cosets of a subgroup $H$ then there is an homomorphism between $G$ and $S_{G:H}$. Why isn't it true if $G$ is not simple? –  Roy Jul 11 '12 at 13:01
    
@DonAntonio: Can we say: "Let $H$" of order $20$ and $K$ of order $3$ be subgroups of $A_5$. Since the set $HK$ has exactly $60$ elements, it should be $A_5$ itself. $H$ and $K$ are solvable groups so $A_5$ would be either. Contradiction!"??? –  B. S. Jul 11 '12 at 13:05
    
@Roy, of course you cannot claim a group of order 60 is isomorphic with a subgroup of order 6...Now, there always exists a homomorphism $\,G\to Sym_X\,$ when we have an action $\,G\times X\to X\,$ , but if $\,G\,$ is simple then this homomorphism is either trivial or $\,1-1\,$ (embedding of $\,G\,$ into $\,S_X\,$), which in our case cannot be. –  DonAntonio Jul 11 '12 at 13:36
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You're welcome, @BabakSorouh . Just remember that $\,HK\cong H\times K\,$ requires, as inner direct product, that both subgroups are normal and with trivial intersection. It is true that if a group is (isomorphic to) a (finite) direct product of solvable groups then it is solvable. –  DonAntonio Jul 11 '12 at 14:10
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@DonAntonio: I guess I need to read some more about this. Thank you very much! –  Roy Jul 11 '12 at 14:52
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A group of order 20 has a normal Sylow 5-subgroup that is cyclic, $C_5$. We know that the Sylow 2-subgroup of $A_5$ is isomorphic to the Klein 4-group (no 4-cycles in $A_5$).

Assume $G$ is a group of order 20 with the above properties: $C_5\lhd G$, and $V_4\le G$. Then conjugation in $G$ gives an action of $V_4$ on $C_5$, and hence a homomorphism $\phi: V_4\rightarrow Aut(C_5)$. We can easily check that the automorphism group of $C_5$ is cyclic of order 4. It is generated by the automorphism $\sigma:x\mapsto x^2$. This implies that $\phi$ cannot be injective, in other words there is an element $g\in V_4$, $g\neq1_G,$ such that $g$ commutes with all the elements of $C_5$. The group $\langle C_5\cup\{g\}\rangle$ is thus cyclic of order $10$.

But the group $A_5$ has no elements of order ten. Therefore it cannot have a subgroup like $G$ either. Therefore it cannot have any subgroups of order $20$.

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I temporarily deleted this, because I didn't want to get in the way of Olivier's effort to make the OP volunteer more information. Anyway, here's an answer that doesn't use the simplicity of $A_5$, which would make the question easy as seen from DonAntonio's answer. Sorry about any eventual confusion caused by the temporary deletion. –  Jyrki Lahtonen Jul 11 '12 at 13:10
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+1 Very beautiful and interesting proof. Still, it uses some arguments that perhaps are as "deep" or "messy" as simplicity of $\,A_5\,$ , which can be proved by more or less simple arguments of 3-cycles or so. –  DonAntonio Jul 11 '12 at 14:03
    
A fair point, @DonAntonio :-) You did a good job explaining to OP how to finalize the use of that idea. +1 has been there from the beginning. –  Jyrki Lahtonen Jul 11 '12 at 16:40
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We use the following fact:

Suppose $N \trianglelefteq G$. Then $N$ contains every element which has finite order coprime $[G:N]$.

Suppose there is a subgroup of order $20$ in $A_5$. This gives us an homomorphism from $A_5$ to $S_3$, the kernel $K$ of which has order $10$ or $20$ by the first isomorphism theorem. The index of $K$ is coprime to $5$, which implies that $K$ contains every element of order $5$ since $K$ is normal. But there are $24$ elements of order $5$ in $A_5$, a contradiction.

This approach can be used to give a nice proof of the fact that $A_5$ is simple, which is done in this paper by Gallian. There is a slight improvement to his proof: normal subgroups of order $2$ are ruled out since they are necessarily central and $A_5$ has trivial center.

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To the contrary, let $G \leq A_5$ such a group. Then intersection of $G$ with any of the $A_4$s in $A_5$ (there are 5 of them) must be $V_4$. This is due to $|G A_4| = \frac{|G| |A_4|}{|G \cap A_4|}$ and Lagrange's theorem. So $G$ contains all double transpositions in $A_5$. But $(12)(34)(34)(15) = (1 5 2)$, a contradiction with Lagrange's theorem.

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your argument seems to rely on $\,G\,A_4=A_5\,$...can you prove this? –  DonAntonio Jul 11 '12 at 15:18
    
Nice. In other words, the only number which is both a multiple of 4 and a divisor of 20 is 4 itself. +1 –  DonAntonio Jul 11 '12 at 15:34
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Your argument is perfectly valid once you accept the simplicity of $A_{5}$. You could also argue this way, and I mention it because it brings out one or two points not mentioned in other answers, though Jyrki's is the closest in spirit. Let us first remark that if $G$ is a finite group, and $P$ is a Sylow $p$-subgroup of $G$, then Sylow's theorem tells us that $[G:N_{G}(P)] \equiv 1$ (mod $p$). From this we may conclude that a group of order $20$ has one Sylow $5$-subgroup, which is necessarily normal.Suppose then that $G = A_{5}$ has a subgroup $H$ of order $20.$ Let $P$ be the unique Sylow $5$-subgroup of $H$, which is also a Sylow $5$-subgroup of $G.$ Now $P \lhd H,$ so that $H \leq N_{G}(P).$ Hence $[G:N_{G}(P)]$ divides $[G:H] =3,$ so that $[G:N_{G}(P)] = 1$ as this is the only divisor of $3$ congruent to $1$ (mod $5$). But this means that $A_{5}$ has a unique Sylow $5$-subgroup, which is certainly not the case, since $\langle (12345) \rangle$ and $\langle (13245) \rangle $ are different Sylow $5$-subgroups of $G.$ It's a useful general fact that if $X$ is any finite group, and $Q$ is a Sylow $q$-subgroup of $X,$ then whenever $M$ is a subgroup of $X$ containing $N_{X}(Q),$ we have $[X:M] \equiv 1$ (mod $q$) (just apply Sylow's theorem in $M$ and in $X$).

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