choose the correct option..?

Question

Let $G$ be a group and $H$ be a normal subgroup of $ G$ such that $H$ is generated by an element $a$ of order $6.$ Let $b ∈ G$.then $bab^{-1}$ is

Choose the correct option :

$a)$$ a$ or $a^2 $

$b)$$ a$ or $a^3$

$c)$$ a $or $a^4$

$d)$$ a $ or$ a^5$

My attempts :i know that$o(a)=o(bab^{-1})=6$. here i find none of the option is correct,,,but my professor say me that some option is correct...as im confused which option is correct .....

any hints/ solution will be appreciated.....

Answer 1

You do not need to go through the orders of all elements.

Note, that

• the image of the generator $a$ must be a generator
• generators are those $a^k$ with $(k,6)= 1$
• $\Rightarrow o(a) = a$ or $o(a) = a^5$

Answer 2

Since $H$ is cyclic of order 6 and $a$ is a generator, just look at the orders of powers of $a$: $a$ and $a^5$ have order 6; $a^2$ and $a^4$ have order 3; $a^3$ has order 2; and $a^6$ has order 1. So the answer is $a$ or $a^5$.

Another way to say this is that $|Aut H|=2$ and the non-identity element inverts $a$.