Determining Why X is a Subgroup of Y

Question

I'm currently taking a course on Intro. to Abstract Algebra and I am a bit confused on what subgroups and subsets are. During class today, for the set $\mathbb{Z}_6$ = {0, 1, 2, 3, 4, 5}, the professor stated that the subgroups are {0, 2, 4}, {0, 3}, and {o}. He wasn't very clear on the explanation, so could anyone explain to me why and how to determine subgroups? Thanks.

Answer 1

You have to specify what binary operation you are working with here. You must first understand the definition of a subgroup.

A subgroup $H \subset G$ of a group $G$ is a set that is closed under the associative binary operator $\cdot_G$ and contains the identity element $e$ of $G$ such that $a\cdot_G e = e\cdot_G a = a$. Further, for each $a \in H$ there exists a unique $a'\in H$ such that $a\cdot_G a' = a'\cdot_G a = e$.

Notice that in each example your professor gave, the identity element is present and that the set containing the identity element is the smallest subgroup of any group.

Answer 2

A subgroup $H$ of a group $(G,*)$ is a subset of $G$ that is also a group with respect to the binary operation $*$.

In your case the group is $\mathbb{Z}_6$ and the binary operation is addition $+$.

You can check that the subsets $\{ 0,2,4 \}$, $\{ 0,3\}$ and $\{ 0\}$ are indeed subgroups. In the first case note in particular that $2+4 = 0$ since we are adding residue classes mod $6$ (it should really be written $[2]_6 + [4]_6 = [6]_6 = [0]_6$).

To see that these are the only subgroups the quickest way would be to note that $\mathbb{Z}_6$ is a cyclic group and so any subgroup of it must be cyclic. Now go through each element in $\mathbb{Z}_6$ and repeatedly add it with itself to generate all subgroups.

Answer 3

A set is any collection of objects. There doesn't have to be any group operation or group structure to them.

The elements of a subgroup must be a subset of the elements of the group, BUT these elements must form a group when using the group operater.

So the group, $\mathbb Z_6$ is the set {0,1,2,3,4,5} combined with the group operator "+".

So say {1, 3,4} is a subset of {0,1,2,3,4,5} but it is not a subgroup because {1,3,4} combined with "+" is not a group. 1 + 1 = 2 is not an element so the group isn't closed. 0, the identity element, is not in set at all and to be a group, 0, the identity element has to be in the set. The inverses of 1 (which is 5) and the inverse of 4 (which is 2) are not in the set at all.

So {1,3,4:+} is simply not a group at all.

So to be a subgroup: The identity element must be included-- So must an inverse for every element that is included-- And the set must be closed.

There are 64 subsets of {0,1,2,3,4,5} and most of them do not form a group.

A group must contain 0. {0:+} is a subgroup because it contains -0=0 and for every 0, 0+0 = 0 is in the group.

If 1, is in the subgroup, then -1 =5 must be in the subgroup. So must 1+1=2, 2+1 =3, 3+1 =4. So if 1 is in a subgroup the subgroup is the entire group $\mathbb Z_6$. There is no other subgroup containing 1.

If 2 is in the subgroup so is -2 = 4. So is 2+2 = 4. {0,2,4} has all the elements necessary to be a group. So {0,2,4:+} is a subgroup.

If 2 is in the subgroup and so is 3 then 3 -2 = 1 must also be in the group and we discussed that above. If 5 is in the subgroup than so is -5 =1.

So {0,2,4:+} is the only subgroup with 2 but not 1.

If 3 is in the subgroup then so is -3 = 3 and 3+3 = 0. So {0,3: +} is a subgroup. It's the only subgroup with 3 that doesn't also have 1. (If it has 2 or -2 =4 or -1 = 5, then it will have 3-2 or 3+4 or -5 = 1.)

If 4 is in the group then -4 = 2 is in the group and we discussed that above.

If 5 is in the group then -5 = 1 is in the group and we discussed that.

So the only subgroups are {0:+},{0,2,4:+}, {0,3:+} and $\mathbb Z_6$.

Of all the 64 subsets these are the only 4 that form groups when combined with the group operator.

Answer 4

In the present case, the law is addition modulo $6$, and the unit element is $0$.

Now, if $H$ is a subgroup of $\mathbf Z_6$, let $a$ the smallest non-zero element of $H$, if any. If $a\in H$, by closedness of a subgroup, $H$ contains $a+a=2a$, $2a+a=3a$, &c. Thus

• if $a=1$, then $H$ also contains $2,3, 4,5$ and $6\cdot 1=0$. So $H=\mathbf Z_6$.
• if $a=2$, $H$ also contains $2\cdot 2=4$, $3\cdot 2=4+2=0$, $4\cdot 2=0+2=2$. So $H=\{0, 2,4\}$.
• if $a=3$, $H$ also contains $2\cdot 3=3+3=0$, $3\cdot 3=2\cdot 3+3=0+3=3$, so $H=\{0,3\}$
• if $a=4$, $H$ contains $4+4=2<4$, so $a$ can't be equal to $4$.
• Similarly, $a$ can't be equal to $5$, since $5+5=4<5$, $3\cdot 5=3$, $4\cdot5=2$, $5\cdot 5=1$.

We see the subgroup generated by $5$ is the same as the whole group, generated by $1$. This is a general fact: the generators of the group $\mathbf Z_n$ are the numbers $a$ $\;(1\le a\le n-1)$ such that $a$ is coprime to $n$.

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