Let |G| = n. Then G is isomorphic to a set of n different nxn matrices with entries 0,1

Question

Let $|G| = n$. Then $G$ is isomorphic to a set of $n$ different $n\times n$ matrices with entries $0,1$. (Operation is matrix multiplication.)

So far I have...

$G = \{g_1, g_2, ... ,g_n\}$ and the set of matrices can be represented this way. Let $e$ be a column vector (this is column vector so the numbers are in vertical, sorry I don't know how to do that in latex)

$\overrightarrow{e_{0}}$ = [1,0,0 ...0] (there are n-1 zeros)

$\overrightarrow{e_{1}}$ = [0,1,0 ...0]

$\vdots$

$\overrightarrow{e_{n}}$ = [0,0,0 ...1]

so the matrices are:

$M_0 = [\overrightarrow{e_{0}},\overrightarrow{e_{1}}, ...\overrightarrow{e_{n}}]$

$M_1 = [\overrightarrow{e_{1}},\overrightarrow{e_{2}}, ...\overrightarrow{e_{n}}\overrightarrow{e_{0}}]$

$\vdots$

$M_n = [\overrightarrow{e_{0}},\overrightarrow{e_{1}},\overrightarrow{e_{0}} ...\overrightarrow{e_{n-1}}]$

I see there is a permutation

Theorem/facts that I assume helpful but don't know how to use:

• Every group is isomorphic to a group of permutation
• Let G be a finite group, |G| = n, then G is isomorphic to a subgroup of S_n

I'm not sure how to proceed from here to prove that they are isomorphic

Answer 1

You have the right idea. The permutation matrices essentially permute the basis vectors just as $S_n$ permutes numbers. A bijection between the basis and the set of numbers lets you define an isomorphism of the group of permutation matrices with $S_n$. Then you're reduced it to proving that a group with $n$ elements is isomorphic to a subgroup of $S_n$.