Question

Let G be a group, let H be a normal subgroup. Prove that a is an element of G the order of Ha in G/H is a divisor of order a in G.

Answer 1

Question 1.

Let $G$ be a group, let $H$ be a normal subgroup of $G$. Prove that if $a$ is an element of $G$, then the order of $Ha$ in $G/H$ is a divisor of order of $a$ in $G$.

Solution

Suppose the order of $a$ in $G$ is finite and equals $n\in\mathbb{N}$. So, $a^{n}=1$ in $G$. Then in the factor group $G/H$ we have

$(Ha)^{n}=Ha^{n}=H\cdot 1=H,$

which is the identity of $G/H$. Therefore, the order of $Ha$ in $G/H$, which is the smallest $k\in\mathbb{N}$ such that $(Ha)^{k}=H$, divides $n$, i. e. the order of $a$.

If the order of $a$ is infinite, the order of $Ha$ however can be finite. For example, take $G=\mathbb{Z}$, $H=2\mathbb{Z}$ and $a=2$. Then $Ha=H$, so $Ha$ has the order $1$ in $G/H$, but $na\neq 0$ for all $n\in\mathbb{N}$, $n\neq 0$. ∎