$G = \Bbb Z _2 ^5$, $H = \langle(1,1,1,1,1)\rangle$ What is $G/H$ isomorphic to?

Question

$G = \Bbb Z _2 ^5$, $H = \langle(1,1,1,1,1)\rangle$

What is $G/H$ isomorphic to? I know $H = \langle(1,1,1,1,1)\rangle$ is a cyclic subgroup of $ \Bbb Z _2 ^5$ generated by $\langle(1,1,1,1,1)\rangle$.. my book gives a somewhat similar example using cosets that I don't quite understand. Can I do this using the method involving the kernel, or does that not quite work?

Answer 1

The group $G$ has order $32$ and each of its elements has order $1$ or $2$. Therefore, and since $H$ has order $2$, $G/H$ has order $16$ and each of its elements has order $1$ or $2$. So, $G/H\simeq\mathbb{Z}_2^4$.

Answer 2

There is an isomorphism $G\to G$ that sends $e_i \to e_i-e_{i+1}$ if $i<5$ and $e_5\to e_5$. This isomorphism sends $e_1+\cdots+e_5$ to $e_1$. It follows that $G/H$ is isomorphic to $G/H'$ where $H' = \langle (1,0,0,0,0)\rangle$ and thus, in turn, isomorphic to $\mathbb Z_2^4$. To understan the general situation, I would recommend you look at the Smith normal form of matrices over PIDs

Answer 3

This is isomorhphic to $\Bbb Z_2^4$ you can see it using the first isomorphism theorem. Define $\varphi : \Bbb Z_2^5 \to \Bbb Z_2^4$ by

$$ \varphi(x_1,x_2,x_3,x_4,x_5)=(x_1+x_5,x_2+x_5,x_3+x_5,x_4+x_5) $$

I'll leave it to you to show it's a group homomorphism.

I'll also leave it to you to show $\varphi$ is surjective.

Let's look at $\ker \varphi$. Suppose $(x_1,x_2,x_3,x_4,x_5)\in \ker \varphi$ then

$$ (x_1+x_5,x_2+x_5,x_3+x_5,x_4+x_5)=(0,0,0,0) $$

If $x_5=0$ then $(x_1,x_2,x_3,x_4)=(0,0,0,0)$. so $(x_1,x_2,x_3,x_4,x_5)=(0,0,0,0,0)$

If $x_5 =1 $ then $x_i+1=0$ for $i=1,2,3,4$. That is $x_i=1$ for $i=1,2,3,4$ (remember this is mod 2). so $(x_1,x_2,x_3,x_4,x_5)=(1,1,1,1,1)$

So $\ker \varphi =\left< (1,1,1,1,1) \right>$

So by the first Isomorphism Theorem $$ \Bbb Z_2^5 / \ker\varphi \cong \Bbb Z_2^4 $$

Answer 4

$f=(1,1,1,1,1), e_2, e_3, e_4, e_5$ is a basis for $G$: $$ G = \mathbb Z_2f \oplus \mathbb Z_2 e_2 \oplus \mathbb Z_2 e_3 \oplus \mathbb Z_2 e_4 \oplus \mathbb Z_2 e_5 $$ Clearly, $$ H = \mathbb Z_2 f \oplus 0 \oplus 0 \oplus 0 \oplus 0 $$ Therefore $G/H \cong \mathbb Z_2^4$.