Group Theory Cyclic Generators Proof

70 Views Asked by At

Suppose $a$ is a power of $b$, say $a=b^k$, then $b$ is equal to a power of $a$ if and only if $\langle a\rangle=\langle b\rangle$.

I am not even sure how to really start this. I want to say something about how $b$ has order $k$ meaning it has $k$ elements. But I am not sure if this is the correct direction to be thinking or where to go from this point.

1

There are 1 best solutions below

0
On

If $b$ is equal to a power of $a$ then $b\in\langle a \rangle$. Hence, $\langle b\rangle\subset\langle a\rangle$. Since $a$ is a power of $b$ the reverse inclusion is true. Finally, $\langle a\rangle=\langle b\rangle$.

If $\langle b\rangle=\langle a\rangle$, since $b\in \langle b\rangle$, $b\in \langle a\rangle$ and $b$ is a power of $a$.