About the proof that a simple group of order 60 is isomorphic to A5

Question

I am stuck proving that a simple group $G$ of order $60$ is isomorphic to $A_5$.

In particular:

• I have shown $|Syl_5(G)|=6$ and $|Syl_3(G)|=10$, so there must be $6\cdot(5-1)=24$ elements of order $5$ and $10\cdot(3-1)=20$ elements of order $3$.

• I could further show $|Syl_2(G)|\notin\{1,3\}$.

• Now, where I get stuck: I need to show $|Syl_2(G)|\neq15$ and therefore $|Syl_2(G)|=5$. I know I have to use the fact that there are already $20+24=44$ elements of order coprime to $2$. That leaves only $60-44-1=15$ elements to have order 2 or 4, which somehow has to lead to a contradiction to $|Syl_2(G)|=15$. But the 2-Sylow-groups have order $2^2$ and can thus intersect non-trivially. So how can I argue from here?

This can probably be done in a very elementary way, but I just don't see it... so thanks for any help!

Answer 1

I would avoid the counting argument, since there are indeed groups of order $60$ with $15$ Sylow $2$-subgroups, and all elements of order $2$ are shared by several of these Sylow subgroups at once.

Here is a slightly easier way to do this, using a slightly stronger version of Sylow's theorems. Namely,

Theorem: Let $p$ be a prime, and $p^r$ the highest power of $p$ dividing $G$. For any two Sylow $p$-subgroups $P$ and $Q$ of $G$, suppose we have $p^k\le |P\cap Q|$. Then the number of Sylow $p$-subgroups of $G$ - call it $n_p$ - satisfies $n_p\equiv 1\pmod{p^{r-k}}$.

The proof is exactly the same as the usual proof (using group actions), you just have to pay extra attention to orbit sizes along the way.

Now, if $G$ is a group of order $60$, containing $15$ Sylow $2$-subgroups, then since $15\not\equiv 1\pmod{4}$, the theorem implies the existence of two Sylow $2$-subgroups - call them $P$ and $Q$ - with $|P\cap Q|=2$. But then $P\cap Q$ is normal in both $P$ and $Q$, and thus its normalizer, $N_G(P\cap Q)$, has order divisible by $4$, and size at least $|P|+|Q|-|P\cap Q|=6$. Thus $N_G(P\cap Q)$ has size at least $12$.

The action of $G$ on the right cosets of $N_G(P\cap Q)$ gives a homomorphism from $G$ into $S_5$. If we are assuming $G$ is simple, this becomes an isomorphism with $A_5$.

[Note that this is actually a contradiction, since $A_5$ does not have $15$ Sylow $2$-subgroups, but either way, you are done.]

Answer 2

I'll give another proof based on counting which I found easy to follow:

It is $60=2^2\cdot3\cdot5$. Let $H\leq G$ with $[G:H]=n\not=1$. Then since $G$ is simple we have $\rho:G\hookrightarrow S_n\Rightarrow |G| | n!\Rightarrow 2^2\cdot3\cdot5 |n!\Rightarrow n\ge5 $.

• If $n=5$ then $[S_5:\rho(G)]=2 \Rightarrow \rho(G)\lhd S_5\Rightarrow \rho(G)=A_5 \Rightarrow G\cong A_5$
• We suppose now that every subgroup of $G$ has index $>5$ so $n_p=[G:N_G(P)]>2$ for $p=2,3,5$. We have that:

$n_5|2^2\cdot3$ and $n_5\equiv 1\pmod{5}\Rightarrow n_5=6$

$n_2|3\cdot 5$ and $n_2\equiv 1\pmod{2}\Rightarrow n_2=15$

Let $P,Q$ be two distinct Sylow $2$-subgroups of $G$ with $P\cap Q\not=1$ and let $N=N_G(P\cap Q)$. Since $P,Q$ are abelian subgroups we have $P,Q\subseteq N$. It is also $P,Q\subsetneq N$ since $P\not=Q$. So we have that

$4||N|,4<|N|,|N|| 2^2\cdot 3\cdot 5\Rightarrow |N|=2^2\cdot 3$ or $2^2\cdot 5$ or $2^2\cdot 3\cdot 5$

1. If $|N|=2^2\cdot 3\cdot 5=|G|$ then $1\not= P\cap Q\lhd N=G$ contradiction since $G$ is simple
2. If $|N|=2^2\cdot 5$ then $[G:N]=3$ contradiction
3. If $|N|=2^2\cdot 3$ then $[G:N]=5$ contradiction

So it follows that $P\cap Q=1$ for every pair of Sylow $2$- subgroups of $G$. Hence $|G|>n_5(5-1)+n_2(4-1)=69$ contradiction.

Hence, only possible case $G\cong A_5$

EDIT: As David A. Craven pointed, in the case where we have two distinct Sylow $2-$subgroups $P,Q$ with $P\cap Q\not=1$ then $2||P\cap Q|\Rightarrow \exists x\in P\cap Q$ with $o(x)=2$. Then $P,Q \subset C_G(x)\Rightarrow |C_G(x)|>4\Rightarrow 12\leq |C_G(x)|\Rightarrow [G:C_G(x)]\leq 5\checkmark$

Answer 3

Here is an alternative way of proving the theorem, perhaps a simpler one.

Suppose $G$ is a simple group of order $60$.

1. Prove that $G \subset A_6$. Indeed, any simple group of order $60$ contains $n_5=6$ Sylow 5-subgroups because $n_5\equiv 1\pmod 5$, $n_5|2^2⋅3$ and $n_5\neq1$ (otherwise group G is not simple). So only one value of $n_5=6$ satisfies these conditions. Consider the action of the group $G$ by conjugations on the set of Sylow 5-subgroups. Since all Sylow 5-subgroups are conjugate, this action represents permutations of 6 elements. Thus we can conclude that $G$ is isomorphic to some subgroup of $S_6$, i.e. $G\subset S_6$. Now look at the signature homomorphism $G \to S_6 \to C_2 = \{\pm 1\}$. That must be trivial, otherwise $G$ has a subgroup of index 2 and is not simple. Thus $G \subset A_6$. (The last statement is taken from here )
2. Now consider the action of group $G$ on cosets $A_6∕G$. There are only $6=|A_6 |/|G| =360/60$ cosets. The action of group $G$ leaves the coset $eG$ in place and permutates the other 5 cosets. Any of the subgroups in $A_6$ that acts in this way is isomorphic to $A_5$. Hence $G \subset A_5$ and $|G|=|A_5| \Rightarrow G \simeq A_5$.