5
$\begingroup$

Statement

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.

$\endgroup$
1
$\begingroup$

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, $n_p\neq 3^3$.

$\endgroup$
  • $\begingroup$ 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 '14 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. $\endgroup$ – Joe Johnson 126 Sep 18 '14 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 '14 at 2:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.