Further Maths Help

SS__

Badges: 15

Rep:

#1

Report Thread starter 2 years ago

#1

My AS Further Maths exam is tomorrow. I've revised all the topics that may come up but I need help with one of them: GROUPS!

I have two questions:
What's a proper subgroup and why do we need to find it?
How do you find a proper subgroup from a group?

Thanks

0

reply

NotNotBatman

Badges: 20

Rep:

#2

Report 2 years ago

#2

(Original post by SS__)
My AS Further Maths exam is tomorrow. I've revised all the topics that may come up but I need help with one of them: GROUPS!

I have two questions:
What's a proper subgroup and why do we need to find it?
How do you find a proper subgroup from a group?

Thanks

H<G if $H \subset G$ and H is a group under the same binary operation as G.

0

reply

SS__

Badges: 15

Rep:

#3

Report Thread starter 2 years ago

#3

(Original post by NotNotBatman)
H<G if $H \subset G$ and H is a group under the same binary operation as G.

Could you maybe explain that in words with no notations? Thanks

0

reply

NotNotBatman

Badges: 20

Rep:

#4

Report 2 years ago

#4

(Original post by SS__)
Could you maybe explain that in words with no notations? Thanks

H is a proper subgroup of G if the set on which H is defined is contained within the set on which G is defined, but not equal and the binary operations of each group are the same.

0

reply

SS__

Badges: 15

Rep:

#5

Report Thread starter 2 years ago

#5

(Original post by NotNotBatman)
H is a proper subgroup of G if the set on which H is defined is contained within the set on which G is defined, but not equal and the binary operations of each group are the same.

Thanks for that! Could you help me with this question, it's about applying this rule.

M = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} under the binary operation of multiplication modulo 11.

List all the proper subgroups of M.

Thanks again!

0

reply

RDKGames

Study Forum Helper

Badges: 20

Rep:

#6

Report 2 years ago

#6

(Original post by SS__)
Thanks for that! Could you help me with this question, it's about applying this rule.

M = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} under the binary operation of multiplication modulo 11.

List all the proper subgroups of M.

Thanks again!

Not my strong place but seeing as nobody has come to help, here's something:

The two most basic subgroups of any group are: the group itself, and the group containing only its identity element (?).

Every other subgroup is considered to be proper.

To find all the proper subsets, you can pick out elements one by one out of

, so say you pick

, then generate the following group:
$\langle 3 \rangle = \{ 3^n : n \in \mathbb{Z} \} = \{ 3^0, 3^1, 3^2, 3^3, 3^4 \} = \{ 1, 3, 9, 5, 4 \}$ (modulo 11), or putting it in order, $\{ 1,3,4,5,9 \}$ and note that we can discard higher powers because on modulo 11 they just revent back to what we have right now, hence we do not gain any new elements. Hence this is one of the proper subgroups under the same operation.

Also note that for something like $\langle 2 \rangle$ you will get that it is the exact same group as

(check!), hence you won't get any proper subgroups from it. We would here call

a generator of

.

Also note that the amount of elements in a subgroup must divide the number of elements in the main group. I.e. the amount of elements we have in M is 10, and the amount of elements we have in $\langle 3 \rangle$ is 5. So this is satisfied.
The divisors of 10 are (excluding obvious ones): 2 and 5.
Therefore your proper subgroups will contain either 2 or 5 elements.

An easy way to check whether this will happen to your pick of number, say

, and is to raise it to the power of 2 or 5. If the result is 1 (mod 11), then this element will give you a proper subgroup.
Here, $7^2 = 5 \pmod{11}$ and $7^5 = 10 \pmod{11}$ therefore you do not need to cover $\langle 7 \rangle$ .
Picking 10 gives $10^2 = 1 \pmod{11}$ so this will definitely give you a proper subgroup. Indeed, we have that $\langle 10 \rangle = \{ 1, 10\} \subset M$

Anyway, how to do what you need to do should be explained in your theory as the method may be different.

(?) The identity element may be used as a proper subgroup, but it is trivial. Best to check with your exam board to see if you need to mention this as a group or not.

1

reply

SS__

Badges: 15

Rep:

#7

Report Thread starter 2 years ago

#7

(Original post by RDKGames)
Not my strong place but seeing as nobody has come to help, here's something:

The two most basic subgroups of any group are: the group itself, and the group containing only its identity element (?).

Every other subgroup is considered to be proper.

To find all the proper subsets, you can pick out elements one by one out of

, so say you pick

, then generate the following group:
$\langle 3 \rangle = \{ 3^n : n \in \mathbb{Z} \} = \{ 3^0, 3^1, 3^2, 3^3, 3^4 \} = \{ 1, 3, 9, 5, 4 \}$ (modulo 11), or putting it in order, $\{ 1,3,4,5,9 \}$ and note that we can discard higher powers because on modulo 11 they just revent back to what we have right now, hence we do not gain any new elements. Hence this is one of the proper subgroups under the same operation.

Also note that for something like $\langle 2 \rangle$ you will get that it is the exact same group as

(check!), hence you won't get any proper subgroups from it. We would here call

a generator of

.

Also note that the amount of elements in a subgroup must divide the number of elements in the main group. I.e. the amount of elements we have in M is 10, and the amount of elements we have in $\langle 3 \rangle$ is 5. So this is satisfied.
The divisors of 10 are (excluding obvious ones): 2 and 5.
Therefore your proper subgroups will contain either 2 or 5 elements.

An easy way to check whether this will happen to your pick of number, say

, and is to raise it to the power of 2 or 5. If the result is 1 (mod 11), then this element will give you a proper subgroup.
Here, $7^2 = 5 \pmod{11}$ and $7^5 = 10 \pmod{11}$ therefore you do not need to cover $\langle 7 \rangle$ .
Picking 10 gives $10^2 = 1 \pmod{11}$ so this will definitely give you a proper subgroup. Indeed, we have that $\langle 10 \rangle = \{ 1, 10\} \subset M$

Anyway, how to do what you need to do should be explained in your theory as the method may be different.

(?) The identity element may be used as a proper subgroup, but it is trivial. Best to check with your exam board to see if you need to mention this as a group or not.

Thanks so much for this! My exam is tomorrow and don't know what I would've done otherwise!

0

reply

Quick Reply

Oops, nobody has posted
in the last few hours.

See more of what you like on
The Student Room

Today on TSR

Poll

Do you have the space and resources you need to succeed in home learning?

Watched Threads

Oops, nobody has posted
in the last few hours.

See more of what you like on
The Student Room

TSR Support Team

Get Started

Using TSR

Info

TSR Group

Connect with TSR

Further Maths Help

Not what you're looking for? Try…

Quick Reply

Related discussions

Related articles

Oops, nobody has postedin the last few hours.

Oops, nobody is replying to posts.

See more of what you like onThe Student Room

Today on TSR

Poll

Do you have the space and resources you need to succeed in home learning?

Watched Threads

Spotlight

Oops, nobody has postedin the last few hours.

Oops, nobody is replying to posts.

See more of what you like onThe Student Room

TSR Support Team

Get Started

Using TSR

Info

TSR Group

Connect with TSR

Oops, nobody has posted
in the last few hours.

See more of what you like on
The Student Room

Oops, nobody has posted
in the last few hours.

See more of what you like on
The Student Room