Sorry if this may seem trivial - I just started studying Group Theory. This is the problem:
Prove that $(g,h) \rightarrow hg$ does not define a group action with $g$ acting on $h$. Prove instead that this defines an action of $G^\text{op} $ on $G$.
Here is my attempt to prove the problem:
To show the first part, note that $(g,h) \rightarrow hg$ does not satisfy the criteria $g,(h,x)=(gh),x$ for $x\in X$ where $X$ is the set acted upon. This is because, we get $xhg=xgh$ which is not necessarily true.
The second part is true because an operation $*'$ in $G^\text{op}$ is defined as $h*'g=g*h$. Where $*$ is the operation in $G$. This will then allow the mapping to satisfy the criteria of a group action. We have, $(g,(h,x))=(g, xh)=xhg$ and $((g*'h),x)=((h*g),x)=xhg$ thus the function obeys the associativity criteria. The identity $(e,x)$ is direct.
My problem is that I do not fully comprehend the original problem.
a) I do not know what it means by $hg$, whether it is with respect to the operation $*$ in $G$ or $*'$ in $G^\text{op}$.
b) I am not sure if my understanding of $G^\text{op}$ is correct, and if I have used the operations correctly.
May someone please explain if I have any mistakes?
Thank you so much!
I think $hg$ is with respect to the "normal" operation in $G$.
It looks like your understanding of $G^{\text{op}}$ is correct (set $G$ but with group operation "reversed").
My attack on the actual question would be:
For the first part, the set acted upon is the set of the elements of the group $G$. Now let $g,g',h\in G$. Then for $(g,h)\mapsto hg$ to be an action we need $(gg',h)=(g,(g',h))$, but $(gg',h)=hgg'$ and $(g,(g',h))=(g,hg')=hg'g$. These are not necessarily equal if $G$ isn't abelian.
For the second part, we get: $$(g*'g',h)=(g'g,h)=hg'g$$ and (still): $$(g,(g',h))=(g,hg')=hg'g$$ So $(g*'g',h)=(g,(g',h))$, which is what we want for this to be an action of $G^{\text{op}}$ on $G$.