Whenever $G$ is finite and its automorphismus is cyclic we can already conclude that $G$ is cyclic.
Because as we already saw $G$ is abelian and finite, we can use the fundamental theorem of finitely generated abelian groups and say that wlog
$G=\mathbb{Z}/p^k\mathbb{Z} \times \mathbb{Z}/p^j \mathbb{Z}$. But the automorphismgroup isn't abelian and hence isn't cyclic.
For non finite groups the implication isn't true.
The following is from this link and only slightly reworded.
Let $G$ be the subgroup of the additive group of rational numbers comprising those rational numbers that, when written in reduced form, have denominators that are square-free numbers, i.e., there is no prime number $p$ for which $p^2$ divides the denominator.
Then:
The only non-identity automorphism of is the negation map, so the automorphism group is $\mathbb{Z}/2\mathbb{Z}$, and is hence cyclic.
The group $G$ is not a cyclic group. In fact, it is not even a finitely generated group because any finite subset of can only cover finitely many primes in their denominators. It is, however, a locally cyclic group: any finitely generated subgroup is cyclic.
You look like you've got a good idea for part $1,$ though it's hard to read and understand. Here's how I'd adjust it:
Suppose $G$ is a finite group of order $n,$ so that $G =\{x_1,x_2,\dots,x_n\},$ where the $x_i$ are distinct. After applying the transformation $x\mapsto T(x)x^{-1},$ we get $\{T(x_1)x_1^{-1},T(x_2)x_2^{-1},\dots,T(x_n)x_n^{-1}\}.$ We want to show that this set is equal to $G.$ We know that it is a subset of $G$ since $G$ is a group, and if we can show that its elements are distinct, then it will be equal to $G$ by Pigeonhole principle. On the contrary, suppose that the elements are not all distinct, so that for some $i,j$ with $i\ne j,$ we have $$T(x_i)x_i^{-1}=T(x_j)x_j^{-1}.$$ Then $$T(x_j)^{-1}T(x_i)=x_j^{-1}x_i,$$ which leads to $$T(x_j^{-1}x_i)=x_j^{-1}x_i$$ by homomorphism properties, and so $x_j^{-1}x_i=e$ by the assumed property of $T$. But then $x_i=x_j,$ contradicting the fact that $i,j$ (and so $x_i,x_j$) are distinct.
Basically, it's the same thing you were saying, but with improved formatting, phrasing/spelling. I also removed the undesirable assumption (from the original post) that $G$ was abelian. Moreover, I added the assumption that $G$ had order $n.$ (Do you see why that's necessary?) Also, do you see why the condition $T(x)=x\implies x=e$ was important, here?
Edit: It's worth noting, though, that we can do even better, and can proceed directly instead of by contradiction.
Consider any group $G$ and any automorphism $T:G\to G,$ and define a map $f$ by $f(x):=T(x)x^{-1}$ for all $x\in G.$ I claim that the following statements are equivalent:
- $T(x)=x$ only if $x=e.$
- $f$ is a one-to-one map of $G$ into $G.$
Since $T(x)\in G$ and $x^{-1}\in G$ for all $x\in G$ by automorphism and group properties, then we automatically have that $f$ is a map from $G$ into $G$ regardless of any other assumptions.
Now, if $T$ has the property that $T(x)=x$ only if $x=e,$ then taking any $x,y\in G$ such that $f(x)=f(y),$ we have by definition that $T(x)x^{-1}=T(y)y^{-1},$ so $T(y)^{-1}T(x)=y^{-1}x,$ and so $T\left(y^{-1}x\right)=y^{-1}x$ by homomorphism properties. By assumption, it follows that $y^{-1}x=e,$ so $x=y,$ meaning that $f$ is a one-to-one map on $G$.
On the other hand, if we suppose that $f$ is a one-to-one map on $G,$ then take any $x\in G$ and suppose that $T(x)=x.$ Thus, $T(x)x^{-1}=e,$ or equivalently, $f(x)=e,$ but noting that $f(e):=T(e)e^{-1},$ then $f(e)=ee^{-1}$ by homomorphism properties, so $f(e)=e=f(x).$ Since $f$ was assumed to be one-to-one, it follows that $x=e.$ Thus, $T(x)=x$ only if $x=e.$ $\Box$
As an immediate corollary, the following are equivalent for any finite group $G$ and automorphism $T:G\to G.$
- $T(x)=x$ only if $x=e.$
- $x\mapsto T(x)x^{-1}$ is a one-to-one mapping of $G$ onto $G.$
As for part $2,$ let me give you a hint.
The following are equivalent: $$T^2=id\\\forall x\in G,T\bigl(T(x)\bigr)=x\tag{$\star$}$$
The following are also equivalent: $$G\textrm{ is abelian}\\\forall g,h\in G,gh=hg\\\forall g,h\in G,ghg^{-1}h^{-1}=e\\\forall g,h\in G,(gh)^{-1}=g^{-1}h^{-1}\tag{$\heartsuit$}$$
Now, using part 1, we know that for each $g\in G,$ there is an $x\in G$ such that $g=T(x)x^{-1}$--let's call this by the name $x_g,$ so we know which $g$ it corresponds to.
Applying $(\star)$ and part $1$ then gives us $$T(g)=T\bigl(T(x_g)x_g^{-1}\bigr)=T\bigl(T(x_g)\bigr)T(x_g)^{-1}=x_gT(x_g)^{-1}=\bigl(T(x_g)x_g^{-1}\bigr)^{-1}=g^{-1}$$ for each $g\in G.$ Can you use this together with $(\heartsuit)$ to show that $G$ is abelian?
Best Answer
Use the line of reasoning in Show that a finite group with certain automorphism is abelian by Arturo Magidin [AM] and user9413 [U]: If the only fixed point of $f$ is $e$ then $G$ is abelian.
However, I found the first step to show that showing Claim 1 below [AM] [U], to be nontrivial so I worked out the proof for myself.
So here is my proof of Claim 1: Suppose $x$ and $y$ in $G$ satisfy $x^{-1}f(x) = y^{-1}f(y)$ [AM]. Then applying $f$ to both sides yields:
$f(x^{-1}f(x)) = f(x^{-1})f(f(x)) = f(x^{-1})x = f(y^{-1})y$.
Then this implies $f(y)f(x^{-1}) = yx^{-1}$, which implies $f(yx^{-1}) = yx^{-1}$, which would imply that $yx^{-1}$ is a fixed point of $f$, which, by the assumption that $e$ is the only fixed point, implies $yx^{-1} = e$, which implies $y=x$. Thus $h: x \mapsto x^{-1}f(x)$ is a one-to-one of $G$ onto itself, and [as $G$ is finite], implies that the image of $h$ is $G$ itself, implying Claim 1.