The corollary will based on the Lemma above and Cauchy's Theorem: If is a finite group whose order is divisible by a prime , then contains an element of order .
I have this Lemma: If is a finite abelian group whose order is divisible by a prime , then contains an element of order .
I have shown that for , and order of is , then is an element in of order .
Now, suppose that has order , where . Since is abelian, is a normal subgroup of G and is an abelian group of order . It is easy to see that is divisible by (since is divisible by and ) and is less than .
Afterwards I know I need to show that there is an element in whose order is , but I'm stuck here. Please help me in completing the proof.
I have another corollary: A finite group is a -group is a power of .
I have shown .
For , I used Lagrange's Theorem and shown that every subgroup of has order of power of . How to continue the proof from here?
do it by induction on the order of the group: if the order is 1, there's nothing to prove. so suppose |G| = n with n> 1, and the claim is true for any abelian group of order less than n.
let be a prime divisor of |G|. and choose let and if then will divide |G/H| and obviously |G/H| < |G|. so by induction, G/H has
an element of order p, say . so hence which gives you because why? if then m = kp and and you're done!
there's nothing left to continue! here's the definition of a p-group: a (finite or infinite) group is called a p-group if the order of every element of the group is a power of p.I have another corollary: A finite group is a -group is a power of .
For , I used Lagrange's Theorem and shown that every subgroup of has order of power of . How to continue the proof from here?
Proof of the corollary: do induction again: if G is abelian, we're done. if there exists such that p does not divide then since p divides |G|, we must have
since we're done by induction. if for any then by the class equation, and we're done again because Z(G) is abelian. Q.E.D.
for another proof of Cauchy's theorem see here.
In following the proof by NonCommAlg by the conjugacy class equation the only step that needs work to be proven is that if is a finite abelian group and divides then there is an element of order . Here is another way, though I perfer the one by NonCommAlg, involving the decomposition theorem. You can write where are cyclic groups of prime order. Since divides it means divides some . Therefore, . Let . Then and is cyclic therefore it is generated by an element of order . This shows there is an element of order .