I am currently using Herstein's Abstract Algebra 3rd edition for a course on introductory abstract algebra. I am attempting to solve my problem but I am unclear what notation in a problem means.

If $G=S_3$ and $H=\{i,f\}$, where $f: S \to S$ is defined by $f(x_1)=x_2, f(x_2)=x_1, f(x_3)=x_3$ list all of the right and left cosets of $H$ in $G$.

I thought that I was fairly comfortable with cosets, but I have no idea how to approach this question. I am only really familiar with the notation of $S_3$ as the dihedral group, and I have no clue which elements $i$ and $f$ are supposed to be.

  • 3
    $\begingroup$ The element $i$ is presumably the identity in $S_3$, and $f$ is given in the question. $\endgroup$ – David Feb 21 '18 at 3:44
  • $\begingroup$ $S_3$ as the dihedral group Hm? I think $S_n$ usually refers to the symmetry group on $n$ elements, and $D_n$ (or $D_{2n}$) refers to the dihedral group on $n$ elements, although when $n=3$ the two coincide. $\endgroup$ – rschwieb Feb 21 '18 at 14:25

$S_3$ is to be thought of, as the set of all bijective functions from $\{x_1,x_2,x_3\}$ to itself, with composition the product, and the identity map being the identity $i$.

Now, with this in mind, look at $H = \{i,f\}$, where $f$ is given as above, and seen clearly to belong to $S_3$.

First of all, $H$ is definitely a subgroup : $f^2 = i$, so it is closed under product and contains identity and inverse.

Now, how do we find right cosets in general?

Right cosets are equivalence classes under the relation $a \sim b$ if $a = bh$ for some $h \in H$.

So, given any $a \in S_3$, we (right) multiply every element of $H$ with $a$, to get the coset of $a$. Now, find an element not in this coset, and repeat the process with that, until no more elements are left (This process will terminate, since there are finitely many elements in $S_3$).

To simplify, let us switch to one line notation. If $g(x_1) = a, g(x_2) = b$ and $g(x_3) = c$, where $\{a,b,c\} = \{x_1,x_2,x_3\}$, then we will write $g := abc$. For example, our $f = 213$, because it sends $x_1 \to x_2, x_2 \to x_1$ and $x_3$ to itself.

Now, take $a =i=123$, then we get two elements : $\{123,213\} = H$. This is one coset.

Now, $132$ is not in $H$, so repeat : $132 \times 123 = 132$ and $132 \times 213 = 312$. So one coset is $\{132,312\}$.

Finally, $321$ is not included anywhere above, so we get : $321 \times 123 = 321$ and $321 \times 213 = 231$.

Thus, there are three cosets, we can write them succinctly as : $H$, $132 \times H$ and $321 \times H$. Or you can be more explicit by listing the elements of each coset, as I did above.

  • $\begingroup$ The whole coset procedure done here can be carried out for other finite groups as well as infinite groups with some nice structure.Furthermore, $S_n$ is standard notation for the symmetric group, while $D_n$, or $D_{2n}$ is standard notation for the dihedral group, for which this calculation is simple, once you get how the generators of $D_n / D_{2n}$ work with each other. $\endgroup$ – астон вілла олоф мэллбэрг Feb 22 '18 at 23:17

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