SS__
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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
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NotNotBatman
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(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.
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SS__
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(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
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NotNotBatman
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(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.
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(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!
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RDKGames
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(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 M, so say you pick 3, 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 M (check!), hence you won't get any proper subgroups from it. We would here call 2 a generator of M.

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 7, 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.
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(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 M, so say you pick 3, 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 M (check!), hence you won't get any proper subgroups from it. We would here call 2 a generator of M.

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 7, 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!
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