About the group structure of the quotient $G/H$, where $H$ is a normal subgroup.

Question

My question concerns a Theorem we were told in class.

Theorem. $H \lhd G \Longleftrightarrow G/H$ has the structure of a group with operation

$$g_1Hg_2H=g_1g_2H.$$

We were also given an example :

$\Bbb Z/2\Bbb Z:=\{2\Bbb Z,1+2\Bbb Z\}=\{\{evens\}, \{odds\}\}$

where the group structure is given by:

$$\{even\}+\{even\}=\{even\}$$

$$\{odd\}+\{even\}=\{odd\}$$

$$\{odd\}+\{odd\}=\{even\}$$.

My question basically boils down to , how is the theorem being invoked here ?

Edit: Specifically how does the expression $g_1Hg_2H=g_1g_2H$ translate to any of the group structure equations in the example.

Answer 1

Well, the group operation shall be denoted by "+" now. Translating that into $g_1Hg_2H=g_1g_2H$, we get:

$(g_1 + H) + (g_2 + H) = (g_1+g_2)+H$

where inside the parenthesis we are talking about the original group operation, while the outer "+" symbol denotes the quotient group operation.

Let $H = 2\mathbb{Z}$. Thus if you take $g_1+H$ and $g_2+H$ for the equivalence class of even and odd numbers, then we may write these equivalence classes as: $2+H$ and $1+H$, by choosing a class representative (just like you did with $1+2\mathbb{Z}$ and $2\mathbb{Z}$, but i've chosen a different label). Therefore,

$(g_1 + H) + (g_2 + H) = (2+H)+(1+H) = (2+1)+H = 3+H$ which is the same equivalence class of odd numbers, getting the second equality. Similarly you can check for the rest.

Answer 2

Taking $G=\mathbb{Z}$ and $H=2\mathbb{Z}$ in the Theorem gives the example. Since $G$ is abelian, every subgroup is normal. So the quotient $G/H$ is a group with the operation given. It can be identified with the group cyclic group $C_2$ of order $2$.