Suppose $G$ is a simple group of order $60$, show that $G$ can not have a subgroup isomorphic to $ \frac {\bf Z}{6 \bf Z}$.

Of course, one way to do this is to note that only simple group of order $60$ is $A_5$. So if $G$ has a cyclic subgroup of order $6$ then it must have a element $\sigma$ of order $6$, i.e. in (disjoint) cycle decomposition of $\sigma$ there must be a $3$ cycle and at least $2$ transposition, which is impossible in $A_5$.Hence,we are done.

I'm interested in solving this question without using the fact that $ G \cong A_5$. Here is what I tried:

Suppose $G$ has a subgroup say $H$ isomorphic to $ \frac {\bf Z}{6 \bf Z}$, then consider the natural transitive action $G \times \frac {G}{H} \to \frac {G}{H}$, which gives a homomorphism $\phi \colon G \to S_{10}$. Can some one help me to prove that $\ker \phi$ is nontrivial ?

Is there any other way to solve this question? Any hints/ideas?

  • $\begingroup$ $S_{10}$ seems too big to be useful, but if you can find a transitive action on $5$ or $6$ points you can use the fact that the elements of order $6$ in $S_5$ and $S_6$ are odd permutations (see answers). $\endgroup$ Commented Dec 29, 2015 at 14:13

4 Answers 4


(1) Suppose $H=C_6$ is a cyclic subgroup of order $6$ in $G$ (simple group of order $60$). Let $z$ be the element of order $2$ in $H$, so that $\langle z\rangle$ is subgroup of order $2$ in $H$.

(2) $C_G(z)=$centralizer of $z$ in $G$; it clearly contains $H$. Also, $\langle z\rangle$ is contained in Sylow-$2$ subgroup of $G$, say $P_2$, which has order $4$, hence it is abelian, and therefore, centralizer of $z$ contains this Sylow-$2$ subgroup.

(3) Thus, $C_G(z)$ contains a subgroup of order $3$ (of $H$) as well as Sylow-$2$ subgroup of order $4$; hence $|C_G(z)|\geq 12$.

(4) Then $[G\colon C_G(z)]\leq 5$, and since $G$ is simple group of order $60$, $G$ can not have a subgroup of index $<5$ [prove it]. Hence $|C_G(z)|$ must be $12$.

(5) Thus, $H\subseteq C_G(z)$ and $|C_G(z)|=12$. Then $H$ has index $2$ in $C_G(z)$, so $H\trianglelefteq C_G(z)$. If $P_3$ denotes the (unique, characteristic) subgroup of $H$ of order $3$ then it follows that $P_3$ is normal in $C_G(z)$. [Characteristic subgroup of a normal subgroup is normal].

(6) Thus, $P_3$ is a Sylow-$3$ subgroup of $G$, which is normal in a group of order $12$ namely $C_G(z)$, i.e. index of normalizer of a Sylow-$3$ subgroup is at most $5$; it must by $5$ (by similar reason as in (4)).

(7) So $P_3$ is a Sylow-$3$ subgroup of $G$ with index of its normalizer equal to $5$; this means the number of Sylow-$3$ subgroups must be $5$; this contradicts Sylow's theorem.

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    $\begingroup$ Looks good to me. I first thought that you can simplify the endgame by using the fact that $N_G(P_3)$ is always self-normalizing. But your logic does not need that! $\endgroup$ Commented Dec 29, 2015 at 9:16
  • $\begingroup$ oh yes; thats a good point. Please do not delete your comment. (I had never seen such simple application of $N_G(N_G(P))=N_G(P)$.) $\endgroup$
    – p Groups
    Commented Dec 29, 2015 at 9:28

Starting out with a patchwork of Sylow theory. Surely something simpler is out there :-(

A group $G$ of order $60$ has either $1,4$ or $10$ Sylow $3$-subgroups. If it has only $1$, then that is normal. If it has only $4$, then we have a non-trivial homomorphism $f$ from $G$ to $S_4$ meaning that $\operatorname{ker}(f)$ is a non-trivial normal subgroup of $G$.

So we can assume that there are ten distinct Sylow $3$-subgroups. Let $P$ be one of them. Its normalizer $K=N_G(P)$ has order six. Because $H$ certainly normalizes a Sylow $3$-subgroup w.l.o.g. $H=K$. So any element $x\in G$ of order six generates the normalizer of $\langle x^2\rangle$. Therefore the conjugates of $H$ have intersections of order at most two.

In other words there are $20$ elements of order three and $20$ elements of order six in $G$ - two in each conjugate of $H$. There is no room left for the necessary six Sylow $5$-subgroups that $G$ should have in order to be simple.

  • $\begingroup$ Thank you for your solution,Regards! $\endgroup$ Commented Dec 29, 2015 at 9:26

Suppose $a$ has order $6$ and generates the subgroup $A$ of order $6$.

If $A$ is the only subgroup of order $6$ it is normal, so there must be more than one such subgroup. How many are there? Well the normaliser $N$ of $A$ contains $A$ as a subgroup and is not the whole of $G$. That means that it has order $12$ or $30$ (or, as comment suggests, 6 - see below for this).

A subgroup of order $30$ would have index $2$ and would therefore be normal. So the order must be $12$ and there must be $5$ subgroups of order $6$ which are conjugate to $A$. The action of $G$ by conjugation on these subgroups must be faithful (otherwise the non-trivial kernel would be a normal subgroup) and this gives an injective homomorphism of $G$ into $S_5$.

Now an element of order $6$ in $S_5$ is an odd permutation, so the image of $G$ in $S_5$ contains odd permutations. The even permutations in the image form a normal subgroup of index $2$. So if $G$ has a cyclic subgroup of order $6$ it cannot be simple.

Note the key element here that an element of order $6$ in $S_5$ is an odd permutation. The rest of the argument would apply to the nonabelian group of order $6$ as a subgroup, which has to map into $A_5$ if $G$ is to be simple.

For normaliser of order $6$, there are $10$ subgroups of order $6$ containing $20$ distinct elements of order $6$

Sylow gives us six subgroups of order $5$ (we can't have one - it would be normal) containing $24$ elements of order $5$

We can't have one subgroup of order $3$, nor can we have four (action by conjugation gives homomorphism to S_4 which would have a non-trivial kernel) so ten is the only possibility with $20$ elements of order $3$.

And that takes us over $60$ elements already, so the situation is impossible.

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    $\begingroup$ you may have to consider the case when normalizer of $A$ has order $6$. $\endgroup$
    – p Groups
    Commented Dec 29, 2015 at 9:46
  • $\begingroup$ @pGroups Indeed it should. I've added in a bit - not so elegant though, just counting. $\endgroup$ Commented Dec 29, 2015 at 10:02
  • $\begingroup$ This is OK now. (There are many different ways for it; but, I was asking for any one way, just for completeness of your answer). $\endgroup$
    – p Groups
    Commented Dec 29, 2015 at 10:04
  • $\begingroup$ @pGroups I now have a second answer inspired by your comment - not so obvious, but cleaner, I think. $\endgroup$ Commented Dec 29, 2015 at 10:08
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    $\begingroup$ The trick of finding an odd permutation in the homomorphic image inside $S_5$ is nice! $\endgroup$ Commented Dec 29, 2015 at 10:24

Here is another answer based on a simpler observation than my last, inspired by considering a comment by @pGroups on an omission. It is in the same spirit, but is distinct.

Note that a simple group of order $60$ must have six subgroups of order $5$ permuted transitively by conjugation (Sylow). There can't be one subgroup of order $5$ because it would be normal. This action on the subgroups of order $5$ gives an injective homomorphism into $S_6$.

Now note that the elements of order $6$ in $S_6$ are all odd permutations, so if $G$ has an element of order $6$ its image contains odd permutations. The even permutations in the image form a normal subgroup of index $2$, so $G$ cannot be simple.

  • $\begingroup$ This is also good answer; in $S_6$, elements of order $6$ are odd permutations; but, as you proved, $G$ can be embedded in $S_6$, so $[G,G]=G$ can be embedded in $[S_6,S_6]=A_6$. I was initially thinking of such embedding, but not succeeded. $\endgroup$
    – p Groups
    Commented Dec 29, 2015 at 10:11

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