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I am working my way through Charles Pinter's book: A Book of Abstract Algebra. From recommendations on this site, I found a page/web address on Wisconsin University's Math Department that provides solutions to many (perhaps all) of the abundant exercises that are present in Pinter's book.

One of the proposed solutions to Pinter's exercises is the following generalization:

If the product of n elements of a group is the identity element, it remains so no matter in what order the terms are multiplied.

I take issue with this claim and think it is only valid if the group is abelian. Using a simple 3 element example, consider:

$a\circ b \circ c = e$

Using the definition of inverses (and knowing that inverses commute...by defintion) and the associative law, I generated the following cases that must be true:

$a \circ (b \circ c) =e$ and therefore $(b \circ c) \circ a =e$

$(a \circ b) \circ c=e$ and therefore $c \circ (a \circ b)=e$

However, there are still several permutations of this list of elements that require consideration...for example:

$a \circ (c \circ b) =e$

It seems to me this can only be true if the group is abelian. Therefore, should the solution manual be amended to say:

If the product of n elements of an abelian group is the identity element, it remains so no matter in what order the terms are multiplied.

?

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3 Answers 3

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Since this is a true or false question, it is not that the question is phrased incorrectly, but rather that the answer is that it is false.

Your claim that it can only hold if the group is abelian is not true for all such $a, b, c$, which we can see in any group by $a=b=c=e$ and other less trivial examples. What you need to do to show it is false is to pick a specific nonabelian group and find three specific elements where it doesn't work.

Edit: I just noticed that this was not an exercise, but part of the solution manual. In $S_3$, the elements $(12),(23),(321)$ show the original claim is invalid. In fact, if $abc=e$ and $bc\neq cb$, then it is never true that $acb=e$ because $bc$ is the unique inverse of $a$. By similar reasoning, if the claim holds for $a, b, c$, then all three elements commute pairwise. You can show that the group is abelian if this holds true for all triples such that $abc=e$.

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  • $\begingroup$ Probably off topic, but I wonder what would happen if rather than saying the group as a whole is abelian we instead looked at the subgroup generated by $a,b,c$ $\endgroup$ Commented Oct 17, 2019 at 18:24
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    $\begingroup$ @Kitter Yes, it does imply that it is an abelian subgroup (if all permutations result in the identity, not just $acb$). $\endgroup$ Commented Oct 17, 2019 at 18:37
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    $\begingroup$ @Kitter Yes, you're right. $\endgroup$ Commented Oct 17, 2019 at 18:40
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    $\begingroup$ I think the necessary and sufficient condition is that for a product of $n$ $a_i$ then we just need each $a_i$ to be in the centralizer of each $a_j$? $\endgroup$
    – EgoKilla
    Commented Oct 17, 2019 at 18:50
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    $\begingroup$ ohhhhh, i like that! Cheers~ $\endgroup$
    – S.C.
    Commented Oct 17, 2019 at 19:06
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What I was expecting the statement to say was:

If the product of $n$ elements in the group is the identity, it remains so under any cyclic permutation.

Perhaps this is what “order” is meant in the solution? In any case, this might be a reasonable statement to prove for yourself.

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  • $\begingroup$ Sorry for the poor phrasing of my question...and sloppy usage of "if" / "then" (sometimes I forget that the language needs to be precise on the MATH forum). From reading through the comments, my impression is that "If the group in question is abelian, then the statement is true". Is that correct? $\endgroup$
    – S.C.
    Commented Oct 17, 2019 at 19:03
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    $\begingroup$ @S.Cramer If the group is abelian, then you can make the much stronger statement “given any n elements of an abelian group, their product in any order is always equal.” This is because in an abelian group, order never matters. In particular, yes: if the group is abelian then your statement is true. $\endgroup$ Commented Oct 17, 2019 at 19:20
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Counter examples could be found in $2\times 2$ invertible matrices.

For example $A= \begin{bmatrix}1&2\\2&1\end{bmatrix}$,$B= \begin{bmatrix}3&2\\1&5\end{bmatrix}$ $C=(AB)^{-1}$ then obviously $$ABC=I$$ while $$CBA\ne I$$

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