In a group $G$, $bab^{-1}=a^{49} $. order of $a$ is smallest prime dividing order of $G$. What are the possibilities for $p$?

Question

If $G$ is a group of order $n$. $a \in G$ is such that $o(a) = p$ where $p$ is the smallest prime dividing $n$, and if there exists an element $b$ such that $b$ a $b^{-1}=a^{49} $, what are the possible values of $p$? .... If $G$ is abelian then $a=a^{49} $. that is $a^{48} =e$ the identity, that is $p$ divides 48. Then $p$ is either 2 or 3.

In general what we can see is order of $a$ and order of $a^{49} $ is $p$, since they are conjugates, that is $p$ is coprime to 49. that is $p$ cannot be 7.

I could not proceed further, I was thinking about applying a result that says that any subgroup of $G$ of index $p$ where $p$ is the smallest prime dividing $n$, is normal. Could not figure out how to proceed.

Please help.

Answer 1

I will assume you are talking about a finite group, so $b$ has finite order.

$A=\langle a \rangle$ is a subgroup of $G$ isomorphic to the cyclic group of order $p$, $C_p$. Given the property above, the automorphism $\phi_b$ of $G$ given by conjugation by $b$ ($g \mapsto bgb^{-1}$) can be restricted to this subgroup, since it sends the generator $a$ in a power of $a$.

Now, recall that we know all the automorphisms of a cyclic group of prime order, and in particular we know that $|\operatorname{Aut}(C_p)| = p-1$. So, since this is also a group, the order of $\phi_b$ has to divide $p-1$. Now note that the order of $\phi_b$ has also to divide the order of $b$, since applying $\phi_b$ $k$ times is doing $$g \mapsto b^k g (b^{-1})^k$$

But $p$ was the smallest prime dividing $|G|$, so there cannot be common prime factors between $p-1$ and the order of $b$ (since the order of $b$ has to divide $|G|$). This means that the order of $\phi_b$ as an automorphism of $\langle a \rangle$ is $1$, so $$\phi_b(a)=a^{49}=a$$ from which you correctly deduce that $p=2$ or $p=3$ are the only possibilities.