We can prove that for any finite group of order n, it must be true that the order of any of its elements divides n. However, does there always have to be an element of order d for all divisors d of n?
I tried proving this in general and for Abelian groups, but couldn't quite get there.
>>8148416
Obviously not you cretin, that'd imply all groups are cyclic.
>>8148416
the strongest result in this sense is Cauchy theorem
The group [math] (\mathbb{Z}_2)^3 [\math] is an abelian group of order 8 with no elements of order higher than 2.
>>8148461
Forward slash next time friendo
>>8148416
You'd be interested in the Sylow theorems.
>>8148748
Are the Sylow theorems supposed to imply anything about general divisors of the order of the group? As far as I know it only tells you something about divisors that are maximal prime powers?
>>8148575
The reason you "couldn't quite get there" is because if you actually used your brain for half a second, you'd see that it's trivially false. Oh sorry, have you not learned about cyclic groups yet?
>>8148798
It's a special case of what you were asking, since the general case won't work.
