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I am trying to understand subgroups.I know a given a group G under a binary operation ∗, a subset H of G is called a subgroup of G if H also forms a group under the operation ∗.

So if I where to list 5 subgroups of $\mathbb{Z}$ could I simply say, $\mathbb{Z}_2$, $\mathbb{Z}_4$, $\mathbb{Z}_6$, $\mathbb{Z}_7$, $\mathbb{Z}_9$, are all subgroups of $\mathbb{Z}$?

citadel
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user6259845
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    No,these are quotient groups of Z.they are not subgroups – Kislay Tripathi Oct 09 '17 at 04:36
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    Elements of (for example) $\mathbb{Z}_n$ are not even elements of $\mathbb{Z}$, so $\mathbb{Z}_n$ can never be a subgroup. – B. Pasternak Oct 09 '17 at 04:37
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    Although $Z_k \subsetneq \mathbb Z$ then operation $+{\mathbb Z_9}$ is a different operator then the operation $+{\mathbb Z}$. To be a subgroup they have to be a group under the same operation. – fleablood Oct 09 '17 at 04:55

2 Answers2

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The subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$ with $n\in\mathbb{N}$.

Proof:

Let $H\neq\{0\}$ (else simply $n=0$) and let $n=\min\{m|m\in H, m>0\}$.

Suppose $H\neq n\mathbb{Z}$, then exists $k\in H\setminus n\mathbb{Z}$. Then $k=q\cdot n+r$ with $q,r\in\mathbb{Z}$ and $0<r<n$. This is division with remainder.

Hence $0<r=k-qn\in H$. But $r<n$ in contradiction to the choice of $n$. Therefore $H\setminus n\mathbb{Z}=\emptyset$

Cornman
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"H of G is called a subgroup of G if H also forms a group under the operation ∗"

!!AND!!! to be a SUBgroup (and not just a group with a different operation on a subset of $H$) we also need that if $a\in H, b\in H$ then $ab \in H$ is equal to $ab \in G$.

So $\mathbb Z_2 = \{0,1|+:0+0=0;0+1=1+0=1;1+1=0\}$ is NOT a subgroup of $\mathbb Z$ because in $\mathbb Z$, we must have $1+1 = 2 \ne 0$.

And $\mathbb Z_9$ is not a subgroup because in $5+6=11 \in \mathbb Z$ but $5+6 = 2 \in \mathbb Z_9$.

Instead though $2\mathbb Z = \{2*a|a \in \mathbb Z|+\}$ (i.e. the even integers) is a subgroup.

And $k\mathbb Z = \{k*a|a \in \mathbb Z|+\}$ (the multiples of $k$) is a subgroup.

fleablood
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