Showing that a subset is a group

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In summary, the conversation discusses the proof that Gm, defined as the set of elements in a finite, additive, and abelian group G whose order divides m, is a subgroup of G. The proof involves showing that Gm is closed under addition, which can be achieved by considering the least common multiple (lcm) of the orders of two elements in Gm. The conversation also confirms that the lcm is indeed less than m, making it a valid approach for the proof.
  • #1
erogard
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Hi,

Consider a finite, additive and abelian group [tex]G[/tex] of order [tex]mn[/tex], where [tex] gcd(m,n)=1[/tex]. Basically m an n are relatively prime.

Now we define [tex] G_{m} = [/tex] { [tex] g \in G | [/tex] order(g) divides m }.

I'm trying to show that [tex]G_{m}[/tex] is a subgroup of G. All I really need to show is that it is closed under addition. That is, [tex] g_{1}, g_{2} \in G \Rightarrow g_{1} + g_{2} \in G [/tex].

So I let [tex] k_{1} [/tex] and [tex] k_{2} [/tex] be the order of g1 and g2, respectively, and try to express the order of (g1 + g2) in terms of k1 & k2, and, moreover, to show that order (g1+g2) divides m.

I know that k1k2 does the job, that is, k1k2(g1 + g2) = e, but k1k2 may not divide m since we might end up with k1k2 > m.

I've tried a lot of different things, so far unsuccessfully, but any ideas or suggestions would be much appreciated.
 
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  • #2
How about lcm(k1,k2) (least common multiple). That works doesn't it? It's certainly less than m.
 
  • #3
Dick said:
How about lcm(k1,k2) (least common multiple). That works doesn't it? It's certainly less than m.

Thanks, that certainly does it - I'm just trying to convince myself that it is less than m.
 
  • #4
erogard said:
Thanks, that certainly does it - I'm just trying to convince myself that it is less than m.

Then yourself is hard to convince. m IS a common multiple of k1 and k2.
 
  • #5
Dick said:
Then yourself is hard to convince. m IS a common multiple of k1 and k2.

haha, right, I think I've been a bit slow thinking last night. Thanks for your help.
 

1. What is a group?

A group is a mathematical structure that consists of a set of elements and a binary operation that combines any two elements to produce a third element in the same set. The binary operation must also follow certain properties, such as associativity, identity, and invertibility.

2. How do you show that a subset is a group?

To show that a subset is a group, you must first prove that it satisfies the four group axioms: closure, associativity, identity, and invertibility. This means that the subset must be closed under the given binary operation, the operation must be associative, there must be an identity element in the subset, and every element must have an inverse in the subset.

3. What is the difference between a subgroup and a subset?

A subgroup is a subset of a group that itself forms a group under the same binary operation. This means that a subgroup must satisfy all four group axioms, whereas a subset may not necessarily form a group under the given operation.

4. Can a subset of a non-abelian group be abelian?

Yes, a subset of a non-abelian group can be abelian. This is because the subset may only contain elements that commute with each other under the given binary operation, while the entire group may still have elements that do not commute.

5. How do you prove that a subset is a normal subgroup?

To prove that a subset is a normal subgroup, you must show that it is a subgroup and that it is closed under conjugation by any element in the original group. This means that for any element in the normal subgroup, its conjugate by any other element in the group is also in the normal subgroup.

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