Showing that for $G = \langle x \rangle$ if $|G|$ is odd, then $\phi$ is an injection.

I'm given that $G$ is a finite cyclic group and the map $\phi: G \rightarrow G$ defined by $\phi(g) = g^2$, and asked to show that $\phi$ is a homomorphism which is straightforward but also that $\phi$ is 1-1 $\iff |G|$ is odd. I feel like I'm close but can't finish it off. If we assume that $|G|=2k+1$ for some $k \in \mathbb{Z}$ then for some $a,b\in G$ we have can assume for some $a,b \in G$ $\phi(a)=\phi(b)$ and because $a =x^l$ and $b=x^m$ because they are elements of the cyclic group $G$ then, $$ x^{2l}=x^{2m} $$ But I can't seem to land at $a=b$.

I also can't get the reverse direction, so far I have assumed that $\phi$ is 1-1, and so again for $a,b \in G$ $a^2=b^2 \implies a=b$ with $a =x^l$ and $b=x^m$ we have that $$ x^{2l}=x^{2m} $$ Then $x^{2l-2m}=1$ but here the exponent is even. Any hints would be greatly appreciated, thanks in advance.

Solution 1:

We show the "if" direction. Suppose $|G|$ is a cyclic group of odd order and that there exist $x$ and $y$ in $G$, with $x \not = y$ that satisfy $x^2=y^2$. Then $x^{-1}y$ is not $e$ because every distinct element in a group $G$ has a unique inverse. However, assuming $x^2=y^2$, then $x^{-2}=y^{-2}$, and so the following hold: $$(x^{-1}y)^2 = x^{-2}y^2 = y^{-2}y^2 = e.$$

So this gives $(x^{-1}y)^2=e$. As $x^{-1}y$ is not $e$, this implies $\langle x^{-1}y \rangle$ a subgroup of $G$ of order exactly $2$. By the fact that $|G|$ is odd, this is impossible by Lagrange's Thm.

So see the only if direction, suppose $|G| =\langle y \rangle$ has order $2k$. Then as $y$ generates $G$, it follows that $y^k$ is not $e$, lest $G$ has $k$ or fewer elements. But what about $(y^k)^2$ though.

Meanwhile, $G$ can be any group and that $\phi$ is an injection still holds.

Suppose that $G$ has odd order and $y^2=x^2$. We first show that $y$ and $x$ commute. Then letting $\alpha$ be the element such that $x \in \langle \alpha \rangle$, note that $x^2$ and thus $y^2$ is in $\langle \alpha \rangle$. Let $q$ be the order of $\langle \alpha \rangle$, then $q$ is odd, and $y^{2 \left \lceil \frac{q}{2} \right\rceil} = y^{q+1} = y^qy = y$. So , as $y^2$ is in $\langle \alpha \rangle$ and $(y^2)k = y$ for some positive integer $k$, it follows that $y$ is in $\langle \alpha \rangle$ too. Since $x$ and $y$ are in the same cyclic group namely $\langle \alpha \rangle$ they must commute. So here as well $x^{-1}y$ is not $e$ and $(x^{-1}y)^2= x^{-2}y^2$. Finish as above.

Meanwhile, if $G$ has even order, then whether $G$ is abelian or not, there is an element $y \in G$ such that $y \not = e$ yet $y^2=e$. If G is a group of even order, prove it has an element $a \neq e$ satisfying $a^2 = e$. . Again, finish as above.

Solution 2:

Hints:

You have $x^{2l}=x^{2m}$, where $x$ has order $2k+1$. Try raising both sides to the power of $k$.

For the other direction, if $x$ has order $2k$, can you write down two things which square to the identity?