Show that there are no simple groups of order $27p$ for any $p$ prime number.

I got stuck with this problem, I'll write what I've done so far:

Suppose $G$ is a group with $|G|=3^3p$, $p$ prime.

We trivially have that the statement is true if $G$ is abelian, so suppose $G$ is not abelian.

I consider two cases separately:

(1) $p=3$

(2) $p \neq 3$

If we are in case (1), then $G=3^4$, but any group $G$ of the form $G=p^n$ with $p$ prime satisfies $Z(G) \neq \{1_G\}$, as $G$ is not abelian, we have that the center of $G$, $Z(G)$, is a proper normal subgroup of $G$ (it is easy to show normality).

For case (2), if $n_p$ is the number of $p-$ Sylow subgroups and $n_3$ is the number of $3-$Sylow subgroups, then applying Sylow theorems we have the conditions:$$(i) \space n_p \equiv 1 (p), \space n_p|3^3 \implies n_p \in \{1,3,3^2,3^3\}$$ $$(ii) n_3 \equiv 1 (3), \space n_3|p \implies n_3 \in \{1,p\}.$$

If $n_p$ or $n_3$ is $1$, then there is one normal subgroup in $G$, so suppose $n_p,n_3 \neq 1$. Then, we have $n_3=p$ and $n_p \in \{3,3^2,3^3\}$.

-If $n_p$ is $3$, then $3=n_p\equiv 1 (p)$, so we would have $3=1+pk>p$, on the other hand, $p=n_3\equiv 1 (3)$ which means $p=1+3j>3$, which is clearly absurd.

-If $n_p=3^2$, by the same argument of congruence modulo $p$, we have $3^2-1=pk$, but then $p|3-1=2$ or $p|3+1=4$. If $p|2$, then $p=2$, which is absurd by the condition $n_3=p$. If $n_p|4$, then $3<p\leq 4$, there is no prime greater than $3$ and less than or equal to $4$. In conclusion, we can rule out the case $n_p=3^2$.

I don't know what to do for the case $n_p=3^3$. I know that for a simple group $G$ and $H$ a proper subgroup of $G$ of index $r$, $G|r!$, but I have no idea if this result could be useful for this exercise.

I would appreciate suggestions or an alternative/completed solution. Thanks in advance.


1 Answer 1


Any two distinct subgroups of order $p$ must intersect trivially, since they are cyclic. So, if you had $27$ subgroups, that would give you at least $27(p-1)+1$ elements in $G$. The fact that $n_3=p$ gives you at least $26+p-1$ elements for a total of $27(p-1)+26+p$ elements. A little arithmetic: $$ 27p-27+26+p=28p-1 $$ elements, which is too many. So, $n_p\neq 3^3$.

  • $\begingroup$ The following argument is not so clear to me, or at least I don't understand it:"The fact that $n_3=p$ gives you at least $26+p-1$ elements...", There is at least one $3-$ Sylow subgroup, so I have at least $27(p-1)+27=27p$ elements in $G$, but I don't see how you get to "at least $26+p-1$". $\endgroup$
    – user16924
    Sep 18, 2014 at 2:13
  • 1
    $\begingroup$ @user16924 One of the $3$-Sylow groups gives you $26$ elements (without the identity). Each other $3$-Sylow has to give you at least one more element. $\endgroup$
    – J126
    Sep 18, 2014 at 2:43
  • $\begingroup$ Thanks!, sorry I couldn't see it. I am going to accept your answer, if you have an alternative solution using the result:"$G$ simple group, then if $H$ is a proper subgroup of index $r$, $|G|$ divides $r!$, you are welcome to add it to your previous answer. $\endgroup$
    – user16924
    Sep 18, 2014 at 2:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .