Subgroups of the additive group of the remainders mod 8 $(\mathbb{Z}_8, \underline{+}$

Question

Identify the subgroups of the additive group of the remainders mod 8 $(\mathbb{Z}_8, \underline{+})$ where $\mathbb{Z} = \{0,1,2,...,7\}$ and the operation $\underline{+}$ is defined as:

$$ a \underline{+} b = \begin{cases}a+b, &a+b < 8 \\ a+b-8, &a+b \geq 8 \end{cases} $$

What are the subgroups of $(\mathbb{Z}_8, \underline{+})$ and which of them are normal subgroups?

From definition, I know that since $(\mathbb{Z}_8, \underline{+})$ contains $8$ elements then the subgroups can contain either $1, 2, 4 \text{ or } 8$ elements. But I do determine the groups and it's elements?

Answer 1

So a group with $1$ element will be $\{0\}$. That was, literally, trivial.

A group with $2$ elements will be $\{0,a\}$ so that $a + a = 0; a\ne 0$ so of all $2*a \equiv 0 \mod 8\implies 2a = 8k \implies a =4k \implies a\equiv 0 \mod 4$ so $a = 0, 4$ but $a\ne 0$ so the group with $2$ elements is $\{0,4\}$.

Now after doing the above it "should be obvious to even a dope" that $\{0,2,4,6\}$ is a group with four elements. But perhaps less insultingly we know that up to homeomorphism $\mathbb Z_4$ and $\mathbb Z_2 + \mathbb Z_2 $ are the only groups with four elements. So if we look at the individual elements...

$|0| = 0; |1| =8; |2| = 4; |3| = 8; |4| =2; |5| = 8; |6|= 4; |7| = 8$.

So ... scratch everything and start again.

The only group with $1$ element is $<e> = \{e\} = \{0\}$.

The only group with $2$ elements is $<a>$~$\mathbb Z_2$ where $|a| = 2$. The only $a; |a| =2$ is $a =4$. So the only subgroup with $2$ elements is $<4> = \{4\}$.

The only groups with $4$ elements are $<a>$~$\mathbb Z_4$ where $|a| = 4$ or $<a,b>$~$\mathbb Z_2 \times \mathbb Z_2$. where $|a|=2; |b|= 2; a\ne b$.

We have $|2|,|6| = 4$ and so $<2> = <6> = \{0,2,4,6\}$ is one subgroup.

We only have $|4| =2$ so we don't have any $|b| = 2; b\ne a$ so we have no other subgroups.

And the only subgroup with $8$ elements is the group itself.

Answer 2

$\Bbb Z_8$ is a cyclic group. Subgroups of a cyclic group are cyclic and to each positive divisor of the order of the group (here, 8) there corresponds a unique subgroup of that order.

Look at the elements of $\Bbb Z_8$ and the cyclic subgroups generated by them. This will indeed give you all the subgroups, but some of them will be the same subgroup.

Two elements will generate the same subgroup iff their order is the same. This will help you recognize which of them are the same subgroups.

(As mentioned in the comments, all the subgroups will be normal because the group $\Bbb Z_8$ is Abelian)