# Can $G$ have a Sylow -5 subgroups and Sylow -3 subgroups [CSIR-NET-DEC-2015]

Let $G$ be a simple group of order $60$. Then

1. $G$ has six Sylow -5 subgroups.
2. $G$ has four Sylow -3 subgroups.
3. $G$ has a cyclic subgroup of order 6.
4. $G$ has a unique element of order $2$.

$60=2^2.3.5$ No. of Sylow -5 subgroups =$1+5k$ divides 12.So $1+5k=1,6\implies n_5=1,6\implies n_5=6$ as $G$ is a simple group.

Consider $n_3=1+3k$ divides $20\implies 1+3k=1,4,10\implies 1+3k=4,10$. If $n_3=4$ then we have $8$ elements of order $3$ and $A_5$ has 20 elements of order $3$ which is a contradiction.Hence $n_3=10$.

Since $A_5$ has no element of order $6$.So 3 is false.

$A_5$ has many elements of order $2$ viz. $(12)(34),(13)(24),,$. Hence $1$ is correct only .Please can someone check whether I am correct /not?

• The element $(12)(34) \in A_5$ has order $2$. – Travis Jan 5 '16 at 4:34
• Are the other options correct @Travis – Learnmore Jan 5 '16 at 4:51
• Like you say, (1) is true, (2) is false, (3) is false, and (4) is false, but from the wording I'd guess you're expected to draw these conclusions only using simplicity and not referring the properties of the group $A_5$, as it takes some work to establish that this is the only simple group of order $60$. – Travis Jan 5 '16 at 4:58
• thank you @Travis; I will definitely try that – Learnmore Jan 5 '16 at 5:02
• A hint regarding statement (2): To show that $G$ must have more than $4$ Sylow-$3$ subgroups one normally looks at the operation of $G$ on $Syl_3(G)$ ... does that help you already? – jpvee Jan 5 '16 at 9:39

As was remarked before, you do not have to assume that $G \cong A_5$.

(1) You did that one correctly!

(2) If $|Syl_3(G)|=4$, and $S \in Syl_3(G)$, then $|G:N_G(S)|=4$. Now let $G$ act on the left cosets of $N_G(S)$ by left multiplication, then the kernel of this action is $core_G(N_G(S))=\bigcap_{g \in G}N_G(S)^g$, which is a normal subgroup. Hence, by the simplicity of $G$, it must be trivial and $G$ can embedded in $S_4$, a contradiction, since $60 \nmid 24$.

(3) We prove that if a non-abelian simple $G$, with $|G|=60$, has an abelian subgroup of order $6$, then $G \cong A_5$. This gives a contradiction, since it is easily seen that $A_5$ does not contain any elements of order $6$ (note that an abelian group of order $6$ must be cyclic).
So assume $H \lt G$ is abelian and $|H|=6$. $H$ is not normal so, $N_G(H)$ is a proper subgroup (if not then $H$ would be normal) and since $|G:N_G(H)| \mid 10$, we must have $|G:N_G(H)|=5$ ($=2$ is not possible since subgroups of index $2$ are normal). Similarly as in (2), $G/core_G(N_G(H))$ embeds homomorphically in $S_5$ this time. Of course $core_G(N_G(H))=1$, so $G$ is isomorphic to a subgroup of $S_5$ and since it is simple it must be isomorphic to $A_5$ (if we write also $G$ for the image in $S_5$, consider $G \cap A_5 \lhd G$ and use $|S_5:A_5|=2$).

(4) In general: if $G$ is a group with a unique element $x$ of order $2$, then $x \in Z(G)$. Why? Because for every $g \in G$, $g^{-1}xg$ has also order $2$ and must be equal to $x$. In your case $G$ is non-abelian simple, so $Z(G)=1$.

So only (1) is the true statement.

Edit For case (3) I forgot the case where $|G:N_G(H)|=10$. I have a proof that is quite sophisticated and maybe there is an easier way.

Anyway, in this case $H=N_G(H)$. Consider the subgroup $P$ of order $3$ of $H$. This must be a Sylow $3$-subgroup of $G$, since $3$ is the higest power of $3$ dividing $|G|=60$. Observe that in fact $N_G(P)=H$. This follows from what we showed in (2): $|Syl_3(G)|=|G:N_G(P)|=10$ and of course $H \subseteq N_G(P)$.

Trivially, $P \subset Z(N_G(P))$. Now $P$ satifies the criterion of Burnside's Normal $p$-Complement Theorem, see for example Theorem (5.13) here. But then $P$ has a normal complement $N$, such that $G=PN$ and $P \cap N=1$. Now $G$ is non-abelian simple, so $N=1$ or $N=G$, which both lead to a contradiction.