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I understand that we can sometimes shorten proofs when we get something like $a\cdot b\cdot c$ in a associativity proof from one side of the equation.

However, are there some pitfalls to fall in when using the symmetry argument?

I lack the experience, so maybe you can illuminate me.

An example where I would use symmetry is the following:

Let $A$ be a non empty set and $R$ a ring. Let $map(A,R)$ be all functions from $A$ to $R$. Show that the addition defined by $(f+g)(x)=f(x)+g(x)$ for all $x\in A$ is associative.

$((f+g)+h)(x)=(f+g)(x)+h(x)=f(x)+g(x)+h(x)$

by symmetry we get that this addition is associative

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    IMHO this would be more clear, and you would probably get more useful answers, if you gave a specific example of the sort of argument you mean. – 2017-01-16
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    @David ok, I'll add one – 2017-01-16
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    It is rather unsatisfactory to say "by symmetry": symmetry is sometimes a synonym with commutativity. Rather, one should say that, since addition is commutative, and since one can check commutativity *pointwise*, addition so defined is commutative. Sometimes the phrase"by symmetry" is used when one wants to prove something for certain "configuration of arguments", and one can check that the hypothesis on such configuration are stable by symmetries, and thus it suffices to prove the claim for a configuration of choice. – 2017-01-16
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    Your proof, to be convincing and clear, should explicitly invoke the associativity on $+$ in $R$. That is, $(f+g)(x) + h(x) = (f(x)+g(x))+h(x) = f(x)+(g(x)+h(x))$, where the second equality follows from the associativity of $+$ in $R$. From this point onward, you apply the definition of $+$ in $A$ to get to $(f + (g+h))(x)$. – 2017-01-16
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    Your question, "when can one use a blah argument?" has a simple answer. Always. There are no limitations whatsoever as to the sort of arguments you can make. Of course, the arguments you make have to be correct... – 2017-01-16
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    @MarianoSuárez-Álvarez thanks – 2017-01-16

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