Group Theory: G/H is cyclic = for some a in G I want to show( I don't know if its true) Given n in Z, Ha^n = H(a^(-1))^k for some k in Z

Question

Group Theory:
G/H is cyclic = <Ha> for some a in G
I want to show( I don't know if its true)
Given n in Z, Ha^n = H(a^(-1))^k for some k in Z

anonymous · Answer

Since $Ha^{-1} \in G/H$ and $Ha$ generates $G/H$, it is clear that there exists a minimal value $m \in \mathbb N$ such that $Ha^m = Ha^{-1}$.  Raising both sides to the power $n$ gives $(Ha^n)^m = Ha^{-n}$, so $Ha^n = Ha^{-n/m}$, so your statement is true if and only if $n$ is an integer multiple of $m$ (specifically $n=km$).

zzr0ck3r · Answer

ty

zzr0ck3r · Answer

@yakeyglee I have a quick question
If group G has finite order then, does every element in G form a cyclic subgroup?

anonymous · Answer

Rather, every element in $G$ generates a cyclic subgroup, yes.  This, in tandem with Lagrange's theorem, tells us that the order of an element in a group divides the order of the group.

zzr0ck3r · Answer

ty