Let $G$ be a group of order order 10, such that $a,b\in G$. $|a|=5$ and $|b|=2$. Prove $bab^{-1}$ is equal to $a$ or $a^{-1}$. I'm studying for an algebra exam by doing past papers and I've currently got stuck on the following problem, which I've been trying to solve for some time:
Let $G$ be a group of order order 10, such that $a,b\in G$. $|a|=5$ and $|b|=2$. Prove $bab^{-1}$ is equal to $a$ or $a^{-1}$.
I've been given hind to consider the element $b^{2}ab^{-2}$, but I'm still not getting very far.
Any help would be greatly appreciated, thank you 
 A: Since $a$ has order 5 and $G$ is of order 10, the subgroup generated by $a$ is of index 2 in $G$ and hence normal. Thus $bab^{-1}$ is in the subgroup generated by $a$ and hence is a power of $a$. Let $bab^{-1} = a^i$, $i = 1, 2, 3, 4$. Now, $a = b^2ab^{-2} = ba^ib^{-1} = (bab^{-1})^i = (a^i)^i$. When $i = 1, 4$ we have nothing to prove. If $i = 2$, then we get $a^4 = a$, a contradiction. Again if $i=3$, we get $a^9 = a$, again an impossibility. Thus $bab^{-1} = a$ or $a^4 = a^{-1}$
A: With just a bit more work, we can show a more general result: if $p$ is prime, and $|G| = 2p$, and $a,b\in G$ with $|a| = p$ and $|b|=2$, then $bab^{-1}$ is either $a$ or $a^{-1}$.
First note that $\langle a \rangle$ has index $2$ and is therefore normal. This means that $bab^{-1} \in \langle a \rangle$, say $bab^{-1} = a^j$.
Let $\phi$ denote conjugation by $b$, so $\phi(a) = bab^{-1} = a^j$. Then $\phi$ is an automorphism of $\langle a \rangle$, and since $|b| = 2$, we see that $\phi$ has order $2$, i.e. $\phi \circ \phi$ is the identity map on $\langle a\rangle$. So, on one hand $\phi(\phi(a)) = a$, but on the other hand, $\phi$ is a homomorphism, so $\phi(\phi(a)) = \phi(a^j) = (\phi(a))^j = (a^j)^j = a^{j^2}$. Combining these two results, we must have $a^{j^2} = a$.
Thus, modulo $p$, we must have $j^2 = 1$, equivalently $j^2 - 1 = 0$, equivalently $(j-1)(j+1) = 0$. As $\mathbb Z_p$ is a field, this forces either $j-1=0$ or $j+1 = 0$, hence either $j = 1$ or $j = -1$. This means that either $\phi(a) = a$ or $\phi(a) = a^{-1}$.
In either case, we have $G = \langle a \rangle \langle b \rangle$, with $\langle a \rangle \lhd G$ and $\langle a \rangle \cap \langle b \rangle = 1$, so this is a semidirect product of the form $\langle a \rangle \rtimes \langle b \rangle$.
In the case $\phi(a) = a$, we have $ab = ba$, so $a$ and $b$ commute, which means the product is direct: $G = \langle a \rangle \times \langle b \rangle$. Since both factors are abelian, so is $G$.  In fact, since $|a|$ and $|b|$ are relatively prime, $G$ is cyclic.
In the case $\phi(a) = a^{-1}$, we have $ba = a^{-1}b$, which yields the dihedral group of order $2p$.
