- #1

mathmari

Gold Member

MHB

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We have a cyclic group $G$ and its generator $a$. The subgroup $H=\langle a^8\rangle $ of $G$ is also given. Let $b$ be the class of $a$ in $G/H$.

Why is the order of $b$ equal to the order of $G/H$ ? We have that $G/H = \{gH \mid g \in G\}$, or not?

Since $a$ is the generator of $G$ we get that $g=a^i, \ i\in \mathbb{N}$. Then we have that we get $G/H = \{a^{8+i} : i\in \mathbb{N}\rangle\}$, or not?

Is everything correct so far? How could we continue? (Wondering)