Show that $\{[0],[2],[4],[6],[8],[10]\}$ is subgroup of $\Bbb Z_{12}.$

Question

I have set $H=\{[0],[2],[4],[6],[8],[10]\}$ and $G=\mathbb Z_{12}$

I will show that H is supgroup of G

Clearly, $H\neq\emptyset$ and $H\subseteq G$

I can write $H=\{m[2] : m \in \mathbb Z\}$

1. I will show that H is closure, let $m_1[2],m_2[2]\in H$,

I've $m_1[2]+m_2[2]=(m_1+m_2)[2]\in H$

2. I will show that $m_1[2]+(m_2[2])^{-1}\in H$, let $m_1[2],m_2[2]\in H$

I've $m_1[2]+(12-m_2)[2]=(m_1-m_2+12)[2]\in H$

Thus, $H$ is subgroup of $G$

Please check my proof, Thank you.

Answer 1

Alternative proof:

1. $f: \mathbb Z \to \mathbb Z_{12}$, $f(x)= x \mod 12$ is a group homomorphism.

2. Your subset is a subgroup of the range of $f$ because your subset is the image of the subgroup $22\mathbb Z$ of the domain of $f$.

3. Finally, image of group homomorphism preserves subgroups.