How would I prove what elements $S_3$ contains, and what its subgroups are?

Question

Given that $S_3$ is a symmetric group of size three, how would I find all elements of it, and all subgroups?

Answer 1

It is usual to write elements of $S_{3}$ in one of two forms. As @DBF explained, $S_{3}$ consists of all bijections on the set $\{1,2,3\}$. Thus for an element $\sigma\in S_{3}$ you need to explain where $\sigma$ maps each of $1,2\text{ and }3$. Suppose we have that

$ \sigma:1\longmapsto 2\\ \sigma:2\longmapsto 1\\ \sigma:3\longmapsto 3 $

Then we would write this as

$ \sigma=\left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 3 \\ \end{array}\right) $

Here the first column above says that $\sigma:1\mapsto 2$, the second column says $\sigma:2\mapsto 1$ and the third column says $\sigma:3\mapsto 3$.

Another way that you could write this is as $\sigma=(1,2)(3)$. This says that sigma maps $1\mapsto2$ and then $2\mapsto 1$, whilst $3\mapsto 3$. This is known as the cycle decomposition of $\sigma$. Don't worry if you find this trickier though. If you are just starting out with abstract algebra it may be best to stick with the first notation. (However, if you want to read more about cycle decomposition, just look in any abstract algebra or group theory textbook.

With the above, you should be able to write out all 6 elements of $S_{3}$.

Now, what do you know about subgroups of a given group? Well they need to be subsets and they need to be groups in their own right. You have four axioms that determine if a non-empty set together with a binary operation (a way of multiplying two elements together - below I've denoted it by $*$) is a group. It must be

1. Closed under the binary operation (so for all $,y$ in your set, both $x*y$ and $y*x$ are in your set).
2. Multiplication must be associative (so for all $x,y,z$ in your set we have the relation $(x*y)*z=x*(y*z)$ - i.e. it doesn't matter where you place the parentheses).
3. There must be an identity element with respect to multiplication (an element $e$ such the $e* x=x=x* e$ for all elements $x$ of your set).
4. Each element of your set must have an inverse with respect to multiplication (so for all $x$ in your set, there exists some $y$ in your set such that $x* y=e=y* x$).

Aside: The identity element of a group is unique, and for each $x$ in your set, the inverse will be unique, hence it is usually denoted by $x^{-1}$.

Now returning to $S_{3}$, the elements of $S_{3}$ are bijections and hence are functions, so our binary operation/multiplication is just composition of functions. Also the identity element $e$ will just be the identity function:

$ e=\left(\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{array} \right) $

For a subset $H\subset S_{3}$ to be a subgroup, by axiom 3 it must contain the identity element $e$. As $S_{3}$ is a group the associativity rule (axiom 2) will automatically hold in any subset of $S_{3}$, so we now just need to check axioms 1 and 4.

I suggest that you try a few examples out to see what is going on - it is the best way to learn. You can check that $\{e\}\subset S_{3}$ satisfies axioms 1 and 4, so it is a subgroup of $S_{3}$. Next take $H_{a}=\{e,\sigma\}$ where $\sigma$ is as defined above. Does this satisfy the axioms 1 and 4? If not, what element can you add so that these axioms are satisfied?

Next take $H_{b}=\{e,\tau\}$ where

$ \tau=\left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & 1 \\ \end{array}\right) $

Does $H_{b}$ satisfy axioms 1 and 4, and if not, what elements of $S_{3}$ do you need to add to satisfy the axioms?

As others have said, there are many tools such as Lagrange's Theorem that can help you with this, but on a small example such as this, it is probably more beneficial for your understanding to try a few examples.

SPOILER

In the above examples, $H_{a}=\{e,\sigma\}$ is a subgroup of $S_{3}$ as it satisfies axioms 1 and 4. For axiom 1, note that $e*e=e$, $e*\sigma=\sigma$, $\sigma*e=\sigma$ and $\sigma*\sigma=e$. From this we can also see that $e^{-1}=e$ and $\sigma^{-1}=\sigma$. In the second example, $H_{b}$ is not a group as $\tau^{2}=\left(\begin{array}{ccc}1 & 2 & 3 \\ 3 & 1 & 2 \end{array}\right)$. However $H_{c}:=\{e,\tau,\tau^{2}\}$ is a subgroup as it is now closed under multiplication (an easy check) and each element has an inverse, namely $e^{-1}=e$, $\tau^{-1}=\tau^{2}$ and $(\tau^{2})^{-1}=\tau$.

Answer 2

Regarding to @David's nice observations indicating any element of $S_3$ can be considered as a on-one function, say $\sigma$, we get the element of $S_3$ as follows:

$$ \{\left(\begin{array}{ccc}1 & 2 & 3 \\ 1 & 2 & 3 \end{array}\right),\left(\begin{array}{ccc}1 & 2 & 3 \\ 1 & 3 & 2 \end{array}\right),\left(\begin{array}{ccc}1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right),\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 1 & 3 \end{array}\right),\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right),\left(\begin{array}{ccc}1 & 2 & 3 \\ 3 & 1 & 2 \end{array}\right)\} $$ which is equivalently be shown as: $$ \{id, (1)(2,3), (2)(1,3), (3)(1,2), (1,2,3), (1,3,2)\} $$

According to Lagrange's theorem, if $H$ wants to be a subgroup of $S_3$ so $|H|$ must devide the order of the group which is $6$. This means that: $$|H|=1,~2,~3,6$$ Since $|(1,2)|=|(2,3)|=|(1,3)|=2$ so $$H_1=\{id\},\\H_2=\{\langle(1,2)\rangle\}=\{id,(1,2)\}\\H_3=\{\langle(1,3)\rangle\}=\{id,(1,3)\}\\H_4=\{\langle(2,3)\rangle\}=\{id,(2,3)\}$$ Also, have a looking at two other cycles $ (1,2,3), (1,3,2)$, we see that $$(1,2,3)^2=(1,3,2),~~(1,2,3)^3=id$$ and this means that $(1,2,3)$ is of order $3$ and $H_5=\langle(1,2,3)\rangle=\{id, (1,2,3),(1,3,2)\}$ Henece, adding $G$, as a subgroup of itself we get $$H_1,H_2,H_3,H_4,H_5,G$$

Here, is a computational approach in GAP:

gap> LoadPackage("sonata");;
     s:=SymmetricGroup(IsPermGroup,3);;
     e:=Subgroups(s)
     for i in [1..6] do Print(e[i],"~",StructureDescription(e[i]),"\n"); od;

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     Group( () )~1
     Group( [ (2,3) ] )~C2
     Group( [ (1,2) ] )~C2
     Group( [ (1,3) ] )~C2
     Group( [ (1,2,3) ] )~C3
     Group( [ (1,3,2), (1,2) ] )~S3