# Is a group with order $3^3\cdot 5\cdot 7$ possible?

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

• Just take the cyclic group, which exists for any $\;n\in\Bbb N\;$ ... – DonAntonio Nov 10 '18 at 15:09
• Okay so then if a group of order 945 is possible, where am I going wrong while calculating the number of elements ? – excalibirr Nov 10 '18 at 15:11
• 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! – DonAntonio Nov 10 '18 at 15:21
• 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. – 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 $$n$$.