MHB Why is the order of b equal to the order of G/H ?

[mathmari]

Hey!

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)

 

[I like Serena]

Hey mathmari!

Hey!

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?

(Nod)

Since $a$ is the generator of $G$ we get that $g=a^i, \ i\in \mathbb{N}$.

Shouldn't that be $g=a^i, \ i\in \mathbb{Z}$? (Wondering)

Then we have that we get $G/H = \{a^{8+i} : i\in \mathbb{N}\rangle\}$, or not?

Since $a^8$ is the generator of $H$ we get that $H=\{a^{8k} \mid k \in \mathbb Z\}$.

Let $g$ be an element of $G$.
If $\overline g\in G/H$, then $\overline g = \{ga^{8k} \mid k \in \mathbb Z\}$.
Therefore $G/H = \{gH \mid g \in G\} = \{\overline{a^i}\mid i\in \mathbb{Z}\}$ where $\overline{a^i}=\{a^{i+8k}\mid k\in\mathbb Z\}$.

It means that $a^i$ is equivalent to $a^{8+i}$ and more generally to $a^{i+8k}$.
So the only distinct elements are $\overline{a^i}$ where $i=0, ..., 7$.
We can write:
$$G/H=\{\overline{a^i}\mid i=0,...,7\}$$
And:
$$\#G/H=8$$
(Thinking)

 

[mathmari]

Shouldn't that be $g=a^i, \ i\in \mathbb{Z}$? (Wondering)
Since $a^8$ is the generator of $H$ we get that $H=\{a^{8k} \mid k \in \mathbb Z\}$.

Let $g$ be an element of $G$.
If $\overline g\in G/H$, then $\overline g = \{ga^{8k} \mid k \in \mathbb Z\}$.
Therefore $G/H = \{gH \mid g \in G\} = \{\overline{a^i}\mid i\in \mathbb{Z}\}$ where $\overline{a^i}=\{a^{i+8k}\mid k\in\mathbb Z\}$.

It means that $a^i$ is equivalent to $a^{8+i}$ and more generally to $a^{i+8k}$.
So the only distinct elements are $\overline{a^i}$ where $i=0, ..., 7$.
We can write:
$$G/H=\{\overline{a^i}\mid i=0,...,7\}$$
And:
$$\#G/H=8$$
(Thinking)

It is given that the order of $G$ is $20$.

Do we not have the following?

$$|H|=|\langle a^8\rangle=\frac{\text{ord}(a)}{(\text{ord}(a), 8)}=\frac{\text{ord}(G)}{(\text{ord}(G), 8)}=\frac{20}{(20, 8)}=\frac{20}{4}=5$$ And so
$$|G/H|=\frac{|G|}{|H|}=\frac{20}{5}=4$$
(Wondering)

 

[I like Serena]

It is given that the order of $G$ is $20$.

Ah, since it was not given in post #1, I assumed that the order of G was infinity.

Do we not have the following?

$$|H|=|\langle a^8\rangle=\frac{\text{ord}(a)}{(\text{ord}(a), 8)}=\frac{\text{ord}(G)}{(\text{ord}(G), 8)}=\frac{20}{(20, 8)}=\frac{20}{4}=5$$ And so
$$|G/H|=\frac{|G|}{|H|}=\frac{20}{5}=4$$

Yes we do.
Moreover: $H=\langle a^8\rangle=\{1, a^8, a^{16}, a^{4}, a^{12}\}=\{a^{4k}\mid k=0,...,4\}$
(Thinking)

 

[mathmari]

Ah, since it was not given in post #1, I assumed that the order of G was infinity.
Yes we do.
Moreover: $H=\langle a^8\rangle=\{1, a^8, a^{16}, a^{4}, a^{12}\}=\{a^{4k}\mid k=0,...,4\}$
(Thinking)

Ah ok!

But why has $b$ the same order? (Wondering)

 

[I like Serena]

Ah ok!

But why has $b$ the same order?

Isn't $b=aH=\{a^{4k+1}\mid k=0,...,4\}$?
And $b^4 = a^4H=1H$? (Wondering)

Moreover, if $a$ is a generator of $G$, isn't it also a generator of $G/H$? (Wondering)

 

[mathmari]

Isn't $b=aH=\{a^{4k+1}\mid k=0,...,4\}$?
And $b^4 = a^4H=1H$? (Wondering)

Moreover, if $a$ is a generator of $G$, isn't it also a generator of $G/H$? (Wondering)

Ah ok! I got it! Thank you! (Smile)