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

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    $\begingroup$ Just take the cyclic group, which exists for any $\;n\in\Bbb N\;$ ... $\endgroup$ – DonAntonio Nov 10 '18 at 15:09
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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. $\endgroup$ – Angina Seng Nov 10 '18 at 15:22

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 $n$.

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