Subgroups of symmetric group

Question

How to show effectively that every finite group is isomorphic to a subgroup of symmetric group

Answer 1

In a comment, you asked specifically for how to embed the Klein group $V$ into a symmetric group, so I will show this, which is an instructive example.

Here is the operation table for $V$:

$$\begin{array}{c|cccc} {\oplus} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 0 & 3 & 2 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 2 & 1 & 0 \\ \end{array} $$

$V$ has 4 elements, so by Cayley's theorem we can embed it into $S_4$. Pretend that each of the rows of the operation table is a permutation of $0,1,2,3$. For example, the third row $2,3,0,1$ is the permutation that takes $0\mapsto 2, 1\mapsto 3, 2\mapsto 0, 3\mapsto 1$, which people sometimes write as

$$\left({0 1 2 3\atop 2 3 0 1}\right).$$

(In cycle notation, this is $(0 2)(1 3)$.)

We now have four permutations, one for each row:

$$ p_0 = \left({0 1 2 3\atop 0 1 2 3}\right) \\ p_1 = \left({0 1 2 3\atop 1 0 3 2}\right) \\ p_2 = \left({0 1 2 3\atop 2 3 0 1}\right) \\ p_3 = \left({0 1 2 3\atop 3 2 1 0}\right) $$

These four permutations are a realization of $V$ as a subgroup of $S_4$.

For each $i,j\in\{0,1,2,3\}$, we have $p_i\cdot p_j = p_{i\oplus j}$, where $\cdot$ is permutation composition and $\oplus$ is the operation of $V$. For example:

$$p_2\cdot p_3 = {\left({0 1 2 3\atop 2 3 0 1}\right)\cdot \left({0 1 2 3\atop 3 2 1 0}\right)\hphantom{\cdot}} = \\ {\left({0 1 2 3\atop 1 0 3 2}\right)} = \\ p_1= p_{2\oplus3} $$

Once you see how this works, you should go and reread CutieKrait's answer, which generalizes this construction: To embed $\langle G, \ast\rangle$ into a symmetric group, you think of the $i$th row of $G$'s operation table as defining a function $f_i$ which takes $j$ to $i\ast j$. This function can be seen as a permutation of the elements of $G$, and so as an element of $S_{|G|}$. Permutations $f_i$ and $f_j$, composed, yield the permutation $f_{i\ast j}$; permutation $f_e$, where $e$ is the identity of $G$, is the identity permutation, and permutation $f_{j^{-1}}$ is the inverse of permutation $f_j$.

Answer 2

What you're referring to is Cayley's Theorem. Effectively, to construct an explicit isomorphism for a specific group with a subgroup of a symmetric group, one must know the details/structure of the specific group, but one can use Cayley's Theorem and its proof as a road map for doing so.

Answer 3

for each $a\in G$ let $$f_a:G\to G$$ $$f_a(x)=xa$$ Show $f_a$ is a bijection that is $f_a$ is in $Sym(G)$, the symmetric group of all bijections of $G$. Then $$\psi:G\to Sym(G)$$ $$\psi(a)=f_a$$ is a one-to-one homomorphism (an embedding).