# Thread: quotient group of GL2(Z/5Z) subgroups.

1. ## quotient group of GL2(Z/5Z) subgroups.

G=2x2 matrixes with a1a1=m, a1a2=b, a2a1=0, a2a2=1. (where integers are viewed as residue classes in Z/5Z)

H=2x2 matrixes with a1a1=1, a1a2=c, a2a1=0, a2a2=1.

Prove that the quotient/factor group G/H is cyclic and of order 4.

I'm confused because I can't see how its of order 4. I think its should be 5. Any help is appreciated!!

2. Originally Posted by platinumpimp68plus1
G=2x2 matrixes with a1a1=m, a1a2=b, a2a1=0, a2a2=1. (where integers are viewed as residue classes in Z/5Z)

H=2x2 matrixes with a1a1=1, a1a2=c, a2a1=0, a2a2=1.

Prove that the quotient/factor group G/H is cyclic and of order 4.

I'm confused because I can't see how its of order 4. I think its should be 5. Any help is appreciated!!
As the title says, I'll assume G and H are subgroups of $GL_2(F)$, where $F=\mathbb{Z}/5\mathbb{Z}$ (m is not zero in this case). Here is the sketch of the proof.

Let ${(\mathbb{Z}/5\mathbb{Z})}^{\times}$ be a multiplicative group of $\mathbb{Z}/5\mathbb{Z}$ excluding 0. Let $\phi:G \longrightarrow {(\mathbb{Z}/5\mathbb{Z})}^{\times}$ be a surjective group homomorphism such that $\phi(A) = det (A)$, where $A \in G$ (verify this is a well-defined group homomorphism). The kernel of $\phi$ is a group $H$. By the first isomorphism theorem, $|G:H | = 4$ (The order of a multiplicative group ${(\mathbb{Z}/5\mathbb{Z})}^{\times}$ is 4).