- #1
Mathdope
- 139
- 0
[SOLVED] General question on cosets
Assume a group G is finite, abelian. Let p^n be a divisor of |G|, p prime, n >1. Let x in G have ord(x) = p (by Cauchy's theorem).
Here's my question. If you now take the factor group of G/<x> you have an element of that factor group that is also of order p. Since the elements of the factor group are cosets of <x> in G, any element has the form <x>g, where g is in G. Thus it would seem that (<x>g)^p = <x>g^p = <x> for some g in G.
What additional fact(s) would make g have order p^2? Is the fact that <x>g has order p in G/<x> enough to guarantee that g has order p^2 in G?
What prevents g^p = e in G?
In particular, how do I know that somewhere in G lies an element of order p^2?
I expect to feel very foolish when I see the answer.
Assume a group G is finite, abelian. Let p^n be a divisor of |G|, p prime, n >1. Let x in G have ord(x) = p (by Cauchy's theorem).
Here's my question. If you now take the factor group of G/<x> you have an element of that factor group that is also of order p. Since the elements of the factor group are cosets of <x> in G, any element has the form <x>g, where g is in G. Thus it would seem that (<x>g)^p = <x>g^p = <x> for some g in G.
What additional fact(s) would make g have order p^2? Is the fact that <x>g has order p in G/<x> enough to guarantee that g has order p^2 in G?
What prevents g^p = e in G?
In particular, how do I know that somewhere in G lies an element of order p^2?
I expect to feel very foolish when I see the answer.