# About a representation of $A_5$ over $\mathbb{F}_2$

Let $V$ be a $5$-dimensional vector space over $\mathbb{F}_2$ with basis $\{e_1,\cdots,e_5\}$. The group $A_5$ acts on this vector by permutation of basis elements. Let $W$ be the subspace of $V$ generated by $$e_1-e_2,\,\,\,\, e_2-e_3,\,\,\,\, e_3-e_4,\,\,\,\, e_4-e_5.$$ Then $A_5$ also acts on this subspace (or $W$ is $A_5$-invariant subspace of $V$).

I wanted to know whether this $4$-dimensional representation is irreducible.

What I know about this is that if we had a field of characteristic not dividing order of $A_5$ then over that field, this $4$-dimensional representation would have been irreducible; but here field has characteristic $2$.

• More generally, if the prime $p$ does not divide $n$, then the $(n-1)$-dimensional deleted permutation module (as it is known) of $A_n$ is irreducible over ${\mathbb F}_p$. If $p$ divides $n$, then it has a $1$-dimensional submodule with irreducible $(n-2)$-dimensional quotient. – Derek Holt Dec 15 '16 at 10:59
• If $p$ divides $n$ then $U=\langle v_1+\cdots + v_n\rangle$ and $W=\langle v_i-v_j:1\leq i<j\leq n\rangle$ (dimension $n-1$) are $\mathbb{F}_pA_n$-invariant; what is the $(n-2)$ invariant subspace? (I mean, can we give basis for it?) – p Groups Dec 15 '16 at 11:53
• There is no $(n-2)$-dimensional invariant subspace. $W/U$ has dimension $n-2$, but $U$ has no complementary subspace in $W$. – Derek Holt Dec 15 '16 at 11:59

I think the representation is irreducible:

every vector in $W$ is of the form $\lambda_1e_1+\cdots + \lambda_5e_5$ with $\lambda_i\in \{0,1\}$ and $\lambda_1+\cdots + \lambda_5=0$.

The condition $\lambda_i\in \{0,1\}$ and $\lambda_1+\cdots + \lambda_5=0$ implies that

(1) either all $\lambda_i$'s are zero (trivial vector);

(2) exactly two $\lambda_i$'s have value $1$;

(3) exactly four $\lambda_i$'s have value $1$.

• In $W$ consider an arbitrary vector $\lambda_1e_1+\cdots + \lambda_5e_5$ with $\lambda_i$'s in type (2), say $e_1+e_2$. Then by action of $A_5$ we can obtain vectors $$e_2+e_3,\,\,\,\, e_3+e_4,\,\,\,\, e_4+e_5.$$ (for example, apply $(123)$ on $e_1+e_2$ to get $e_2+e_3$).

This means if an $A_5$-invariant subspace of $W$ contains $e_1+e_2$, then it contains the basis $\{e_1+e_2,\cdots e_4+e_5\}$, i.e. that invatiant subspace should be $W$.

• In $W$ suppose there is a vector $\lambda_1e_1+\cdots + \lambda_5e_5$ with $\lambda_i$'s in type (3). For example, let it be $$e_1+e_2+e_3+e_4.$$ Then by applying permutations ($5$-cycles) to this vector, we can obtain $$e_1+e_2+e_3+e_4,\,\,\,\, e_2+e_3+e_4+e_5, \,\,\,\,e_3+e_4+e_5+e_1, \,\,\,\,e_4+e_5+e_1+e_2,$$ and these four vectors are independent in $W$, so the $A_5$-invariant subspace of $W$ should be $W$ itself.

Thus the representation $W$ is irreducible.

• Looks good to me! I was about to write basically the same answer. – Eric Wofsey Dec 15 '16 at 5:22
• I had seen that this representatio of $A_5$ is orthogonal in characteristic $2$; but I am unfamiliar with orthogonal groups over finite fields, hence posted the question. Fortunately, this simple way ensures that there is a $4$-dimensional irreducible representation of $A_5$ over $\mathbb{F}_2$. – p Groups Dec 15 '16 at 5:23