3
$\begingroup$

Is the group $|G|=3^3\cdot 5\cdot 7$ possible?

I've been examining it, and it seems like there can't be enough elements in the Sylow-$p$-subgroups.

That is, there can only be $21$ (or $1$) groups of order $5$ (the only number that is equal to $1 + 5k$ and divides $189$). We also know they can't contain a proper subgroup, so their intersection must be trivial. This implies there are $84$ elements of order $5$, at most.

Similarly, the number of $7$-sylow subgroups can only be $1$ or $15$, and again the intersection is trivial. So, at most $90$ elements are of order $7$.

This leaves elements of order $3$, $9$, or $27$. The number of Sylow subgroups is $1$ or $7$, so the most amount elements can be found if there are seven Sylow $3$ groups which intersect non-trivially, which means there are at most $182$ elements of this type.

But $1+182+84+90 \neq 945$.

$\endgroup$
  • 29
    $\begingroup$ Just take the cyclic group, which exists for any $\;n\in\Bbb N\;$ ... $\endgroup$ – DonAntonio Nov 10 '18 at 15:09
  • 1
    $\begingroup$ Okay so then if a group of order 945 is possible, where am I going wrong while calculating the number of elements ? $\endgroup$ – excalibirr Nov 10 '18 at 15:11
  • 6
    $\begingroup$ You forgot to take into account all the possible of elements of order $\;3\cdot5=15\,,\,3^2\cdot5=45\;$ and etc. There are many possibilities! $\endgroup$ – DonAntonio Nov 10 '18 at 15:21
  • 28
    $\begingroup$ Basically, you are forgetting that there can be elements of the group which are outside the Sylow subgroups; elements whose orders are divisible by more than one prime. $\endgroup$ – Lord Shark the Unknown Nov 10 '18 at 15:22
2
$\begingroup$

Consider the cyclic group of order $3^3\cdot 5\cdot 7$.

In fact, for any $n\in\Bbb N$, there exists at least one group, namely the cyclic group of order $n$.

$\endgroup$

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.