Suppose $G$ is a cyclic group and $a,b\in G$. There doesn't exist any $x\in G$ such that $x^2=a$ and does not exist any $y\in G$ such that $y^2=b$

Question

Suppose G is a cyclic group and $a,b \in G$. There doesn't exist any $x \in G$ such that $x^2=a$. Also, there does not exist any $y \in G$ such that $y^2=b$ then

(a) there exists an element $g \in G$ such that $g^2=ab$

(b) there exists an element $g \in G$ such that $g^3=ab$

(c) the smallest exponent $k>1$ such that $g^k=ab$ for some $g \in G$ is $4$

(d) none of the above

My intuition says $a$ and $b$ must be odd powers of generator say $p$. So, we can write $a=p^{2n+1}$ and $b=p^{2m+1}$ so $ab=p^{2n+2m+2}$ which gives us option (a). Please verify if the approach is right.

Answer 1

Your intuition is correct, but it needs proving. Fortunately, that's a simple matter. Letting $z$ be a generator of $G,$ we know that $a=z^k$ for some $k$. If $k$ were even, then we would have $k=2j$ for some integer $j,$ whence $a=z^{2j}=\left(z^j\right)^2,$ contradicting the hypothesis. Hence, for any generator $z$ of $G,$ $a$ is an odd power of $z;$ likewise $b$ is an odd power of $z.$ From this, it does follow (by the reasoning you gave above) that option (a) holds. An immediate corollary to this is that (c) and (d) do not hold.

We can't prove that (b) never holds, nor that it always holds. For example, in the cyclic group of order $4,$ we can let $a$ be a generator and $b=a^{-1},$ in which case $e^3=e=ab.$ On the other hand, if $G$ is the cyclic group of order $6,$ letting $a$ be a generator and $b=a,$ it is simple to see that for any $g\in G,$ we have $g^3\in\left\{e,a^3\right\},$ but $ab=a^2\notin\left\{e,a^3\right\}.$

Answer 2

Your approach is right, but you only proved that the given conditions imply (a). For a complete answer, you also need to show that they do not imply (b) and (c).

It is easy for (c) since, with your notation, $ab = g^2$ for $g = p^{n+m+1}$. It is a little less trivial for (b): after all, one may imagine that there exists an $h \in G$ such that $h^3 = ab$. Thus you have to exhibit a counter-example, which requires a little argument. You can take for $G$ the cyclic group of order $6$, say $G = \{1, c, c^2, c^3, c^4, c^5\}$ with $c^6 = 1$. Taking $a = b = c$, I let you verify that your hypothesis is verified, but there is no $g \in G$ such that $g^3 = ab = c^2$.