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Let $(M,\cdot )$ be a monoid with at least two elements: Prove that:
a) Any function $f:M\rightarrow M$ with $f(e)=e$ is the power function (i.e. $f(x)=x^{n},n\in \mathbb{N},n\geq 1$) if and only if $M$ is a group, isomorphic with $(\mathbb{Z}_{2},+) $.
b) If $M$ is a finite set with at least three elements, then any permutation $f$, of $M$, with $f(e)=e$ is the power function if and only if $M$ is a group, isomorphic with $(\mathbb{Z}_{3},+) $.

I have no idea on how to solve this.

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The if parts are straightforward enough. Every cyclic group is generated by one element, call it $g$, and thus every monoid homomorphism from a cyclic group is determined by where it sends $g$. If the codomain is the same cyclic group, then every element is $g^k$ for some $k$. Thus every monoid endomorphism of cyclic groups is a power function: $f(x) = f(g^k) = f(g)^k = g^{mk} = x^m$ where $x = g^k$ and $f(g) = g^m$. There are only two functions $\mathbb{Z}_2\to\mathbb{Z}_2$ such that $f(e) = e$ and both are monoid homomorphisms. Similarly, there are only two permutations of $\mathbb{Z}_3$ that fix $e$ and both are monoid homomorphisms.

The other way is a bit more interesting. First, the $\mathbb{Z}_2$ case. Let $g$ be an element of $M$ not equal to $e$. Then, by assumption, the function which sends everything but $e$ to $g$ is of the form $f(x) = x^n$ which is to say every element of $M$ other than $e$ satisfies $x^n = g$ for some $n > 0$. We easily have $g^n = g = x^n$ which if $M$ was a group would imply $g = x$ for all $x\neq e$ and we'd be done. To show that $M$ must be a group we consider the constantly $e$ function. Again, by assumption, we have $x^n = e$ for all $x$ for some $n > 0$. In fact, $n$ must be greater than $1$ otherwise $g = e$ contradicting the assumption of at least two elements. Therefore for every $x$, $xx^{n-1} = e$ and so $x^{n-1}$ is an inverse and all elements are invertible.

The same approach can't be used for the $\mathbb{Z}_3$ case since neither of the functions used above were necessarily permutations. Assume $M$ has $n+1$ elements and label the non-unit from $1$ to $n$. Let $\varphi$ cyclicly permute the $n$ non-units and fix $e$. Since $\varphi(g_1) = g_1^n = g_2 \neq g_1$, $n$ must be greater than $1$. $\varphi^n(x) = x^{n^n} = x$ for all $x$ where $\varphi^n$ means the $n$-fold application of $\varphi$. It's also clear that for all $x \neq e$, $x = g_1^k$ for some $k$, but since $n^n > n$, $e$ must also be $g_1^k$ for some $k$, thus $M$ is a cyclic group. Now use the permutation $\psi$ which swaps $g_1$ and $g_1^2$ (which are distinct and distinct from $e$ because there are at least $3$ elements) and fixes all other elements, then $\psi(x) = x^2$ and $\psi(\psi(x)) = x = x^4$ for all $x$. Thus we have a cyclic group of order $3$.

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    In the middle paragraph, how are you getting $g =x$ for all non identity $x$? – 2017-02-06
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    I don't see it from $g^n= g = x^n$ alone, which I think just shows that $g^{n-1}$ is the identity. To go further you'd need that $n$ is coprime to the order of $M$. – 2017-02-06
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    But here's a way to do it, I think: – 2017-02-06
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    assuming $M$ is a group, which you proved it is, it is true for your chosen $f$ that $g^{-n} = f(g^{-1}) = g = f(g) = g^n$ which shows that $g^2$ = identity. This also shows, since we chose $g \not= e$, that $n$ is odd (else we'd have $f(g) = e$). Now suppose there is an element $h \not= g$ in $M$. Let $f'$ be the function sending all non-identity elements to $h$. The exponent $m$ associated with $f'$ will be odd as was $n$ for $f$, and then $g = g^m = f'(g) = h$, showing $h$ and $g$ were not distinct. Thus $|M| = 2$ – 2017-02-06
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    @BadamBaplan You are correct; I did make an unsupported logical leap there. Another similar way to get the result, though there's nothing wrong with your approach, is once the fact that $M$ is a group is established considering $\varphi(g^2) = g^{2n} = g = g^2$ which would implies $g = e$ which is a contradiction, thus $g^2 = e$. This was for arbitrary $g$ so all elements have order at most $2$ which implies $\varphi(x) = x^1$. – 2017-02-06