In Herstein's Algebra, a product operation between any two elements(say $\sigma$ and $\tau$) of a symmetric group($S_3$) is defined as :

$\sigma = \begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix} $

$\tau= \begin{pmatrix} 1 & 2 & 3\\ 2 & 1 & 3 \end{pmatrix} $

$(1)\sigma\tau = (\sigma(1)\tau) = (2)\tau = 1$

Whereas, in other texts (Gross's lecture),

$\sigma\tau(1) = (\sigma\tau(1)) = \sigma(2) = 3$

I am guessing both are valid. But then are we talking about two different groups here depending on different ways of doing product operation. Or is it that the set $S_3$ is a unique group for both kind of product operation ?

So my question basically is, For a set, if it satisfies all groups condition under many binary operations, then are they different groups corresponding to different binary operation or the same group?

  • $\begingroup$ There is an isomorphism between a group, and the group with the same set but the product operation reversed, the isomorphism is given by the inverse map. This provides an isomorphism between the two group structures given above. $\endgroup$ Aug 30, 2020 at 14:23

1 Answer 1


This is just whether you act on the left or on the right. You have composed $\sigma$ first, then $\tau$ on the first line, and on the second line you have done $\tau$ first, then $\sigma$. So of course you obtain a different answer. However, $$\tau\sigma(1)=\tau(\sigma(1))=\tau(2)=1.$$ So you obtain the same answer.


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