# How to determine subgroups and generators

• January 24th 2010, 01:30 AM
richardbautista
How to determine subgroups and generators
Hello folks!
I'm new to this forum and I just want everyone to help me =)

In our Abstract Algebra class, we are to find for the subgroup(s) of this group:

G = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14} (this is mod 15)

I already searched the net and found out that there are trivial and non-trivial subgroups for this group... But still, I don't know how to determine those subgroups.

One more thing is how to get the generators of the group:
G = {0,1,2,3,4,5,6,7,8,9} (this is mod 10)

I don't know what generators are :(

• January 24th 2010, 07:40 AM
Dinkydoe
So I assume we're looking at the $G = \mathbb{Z}/15\mathbb{Z}$ with " $+$" as operation.

Quote:

I don't know what generators are
Generators of a group are elements that "generate" the group, that is, with these elements you can make any other element of the group.

Example: In $G = (\mathbb{Z}/n\mathbb{Z}, +)$ the element "1" is allways a generator. Namely any element in G can be written as "1+1+1..+1" etc. Groups may have more then one generator. If a group can be generated by one element we call the group cyclic. Any additive group $(\mathbb{Z}/n\mathbb{Z}, +)$ is cyclic.

Even though such a group is cyclic, there may exist more then one element in G that can generate the whole group.

Namely all $a\in G$ with gcd $(a,15) = 1$ are generators! Proper subgroups are generated by elements $a\in G$ with gcd $(a, 15)\neq 1$. (why?)

Thus proper subgroups are: $\left\langle 3\right\rangle, \left\langle 6\right\rangle, \left\langle 9 \right\rangle, \left\langle 12 \right\rangle,\left\langle 5 \right\rangle, \left\langle 10\right\rangle$

Now you try the other one: $G = (\mathbb{Z}/10\mathbb{Z}, +)$
• January 24th 2010, 01:21 PM
richardbautista
Thanks for a flash reply :D

I think, there is no generator for the first group I posted...
Because zero (0) is one of the elements... which cannot be generated using one (1)... (Crying)
• January 24th 2010, 01:24 PM
Dinkydoe
Yes it can be generated: $15 = 0$

"0" is the identity element of G

Don't forget we're looking at residu-classes modulo 15. (with addition "+" as operator, so 14 + 1 = 0)
• January 24th 2010, 01:32 PM
richardbautista
Aw! Thanks again... very fast response :D

so it goes like this:
1 = 1 mod 15 = 1
1 + 1 = 2 mod 15 = 2
1 + 1 + 1 = 3 mod 15 = 3
.
.
.
1 + 1 + ... + 1 = 14 mod 15 = 14
1 + 1 + ... + 1 + 1 = 15 mod 15 = 0

Since all elements (0...14) are generated, then 1 is a generator? Thanks :D

About subgroups, I don't know what do <3>, <6>, etc. represent?

How could I write them as subgroup like this: subgroup1 = { numbers }
is it like this
Subgroup1 = { 3 }
Subgroup2 = { 6 }?
• January 24th 2010, 01:58 PM
Dinkydoe
(1)

$\left\langle 3\right\rangle = \left\{0,3,6,9,12\right\}$
$\left\langle 6\right\rangle = \left\{0,6,12,3,9,\right\}$
$\left\langle 9\right\rangle = \left\{0,9,3,12,6,\right\}$
$\left\langle 12\right\rangle = \left\{0,12,9,6,3\right\}$.

Thus $\left\langle 3\right\rangle =
\left\langle 6\right\rangle = \left\langle 9\right\rangle = \left\langle 12\right\rangle$

" $\left\langle a\right\rangle$ " means everything that can be generated by the element " $a$" under the operation "+" in the group.

The same way we show that $\left\langle 5\right\rangle = \left\langle 10\right\rangle$. (a group with 3 elements)

Thus we have 2 proper subgroups in $\mathbb{Z}/15\mathbb{Z}$

And $\mathbb{Z}/15\mathbb{Z} = \left\langle 1\right\rangle = \left\langle 2\right\rangle = \left\langle 4 \right\rangle = \left\langle 7 \right\rangle = \left\langle 8 \right\rangle = \left\langle 11\right\rangle = \left\langle 13 \right\rangle = \left\langle 14 \right\rangle$