# Thread: Corollary 1. |a| divides |G|

1. ## Corollary 1. |a| divides |G|

I just would like to ask for the proof of this Corollary: |a| divides |G|.

In a finite group, the order of each element of the group divides the order of the group.

(The theorem is Lagrange's Theorem: |H| divides |G|.
If G is a finite group and H is a subgroup of G, then |H| divides |G|. Moreover, the number of distinct left (right) cosets of H in G is |G|/|H|.)

Thank you!!

p.s. (sorry!! I'm not in my mind, this must have been posted under Abstract Algebra)

2. If I understand you correctly, you want to prove that the order of an element divides the (finite) order of the group, and you want to use Lagrange's Theorem to do it. Pick an element a in G, and consider the subgroup generated by that element (i.e. {a^m for m an integer}). What's the order of that subgroup?

3. the order is not given.

4. That wasn't the point. He was asking you to calculate it. Since there are only a finite number of elements in G, the set of all powers of a, a, $\displaystyle a^2$, $\displaystyle a^3$,... is finite. That means that, for some i and j, $\displaystyle a^i= a^j$. then $\displaystyle a^{i- j}= 1$. The subgroup generated by the single element, a, has order $\displaystyle a^{i- j}$. Although how that will help prove that the order of any subgroup will divide the order of the group, I don't see.

Instead, do this: Given group G and subgroup H, look at all the "left cosets" of H. For each element x of G, the set xH= {xh |h in H} is a "left coset" of H. Show that the left cosets "partition" G. That is, each member of G is in one and only one left coset of H. Also show that all left cosets have the same size- the size of G (with e= identity element, eH, the left coset containing the identity, is H itself). The size of G is the size of H times the number of left cosets of H: since those are all integers, the size of H must divide the size of G.