In a commutative semigroup, $a_{\phi(1)}a_{\phi(2)}...a_{\phi(n)}=a_{1}a_{2}...a_{n}$ In the book of Clifford and Preston,"The algebraic theory of semigroups" volume I, I am reading this: If $a_{1},a_{2},...,a_{n}$ are elements of a commutative semigroup $S$ and $\phi$ is any permutation of the set ${1,2,...,n}$, then: $a_{\phi(1)}a_{\phi(2)}...a_{\phi(n)}=a_{1}a_{2}...a_{n}$, and it says it can be easily proved by induction? 
Can someone please help me to see this, as I am stuck in this part?
 A: Suppose it is true for all permutations of a product of $n-1$ elements and let $\varphi$ be a permutation of $1,2,\ldots,n$. Let $k$ be the index such that $\varphi(k)=n$. By induction we have
$$a_1a_2\cdots\cdots a_{n-1}=a_{\varphi(1)}\cdots a_{\varphi(k-1)}a_{\varphi(k+1)}\cdots a_{\varphi(n)}$$
since removing the $n$ from $\varphi$ gives a permutation of $1,2,\ldots,n-1$. Thus we have
$$a_1a_2\cdots\cdots a_{n-1}a_n=a_{\varphi(1)}\cdots a_{\varphi(k-1)}a_{\varphi(k+1)}\cdots a_{\varphi(n)}a_n$$
But $(a_{\varphi(k+1)}\cdots a_{\varphi(n)})a_n=a_n(a_{\varphi(k+1)}\cdots a_{\varphi(n)})$ by the fact that any two elements commute, so
$$a_1a_2\cdots\cdots a_n=a_{\varphi(1)}\cdots a_{\varphi(n)}$$
as desired.
A: A simple example is a good place to start for some intuition on why the result is true. Matt's answer already gives a proof so I won't supply another.
Define the permutation $\phi$ by $\phi(1)=2, \phi(2)=3$ and $\phi(3)=1$
Then take an element in your semigroup $a_1a_2a_3$ and apply $\phi$ to each of the indices and you get $a_{\phi(1)}a_{\phi(2)}a_{\phi(3)}=a_2a_3a_1$ but this can be rearranged to get $a_1a_2a_3$
You should see that this argument will work with any number of elements.
