subgroup problem help pls

[mathsfun]

let G be an agelian group such that the operation on G is denoted additively. Show that {a is an element of G| 2a = 0} os a subgroup of G. Compute the subgroup for G =13.

 

 

 

let G be an abelian group. show that the set of all elements of G of finite order forms a subgroup of G.

 

 

in addition to helping with these questions can you pls explain an abelian group, i know it is commutable but what does that actually mean. also what does additively mean? is the main idea to show 1) closed 2) identity 3) inverse to prove it is a subgroup?

 

thanks so much!!

[mathsfun]

let G be an agelian group such that the operation on G is denoted additively. Show that {a is an element of G| 2a = 0} os a subgroup of G. Compute the subgroup for G =13.

 

 

 

let G be an abelian group. show that the set of all elements of G of finite order forms a subgroup of G.

 

 

in addition to helping with these questions can you pls explain an abelian group, i know it is commutable but what does that actually mean. also what does additively mean? is the main idea to show 1) closed 2) identity 3) inverse to prove it is a subgroup?

 

thanks so much!!

[matt grime]

Let * denote the group operation. Commutative/Abelian means x*y=y*x for all x and y.

 

Or if you like when you see an expression of the form:

 

x*y*z*w*p*q

 

(which is well defined because of transitivity) then you can rearrange the terms to get it into a better form.

 

Because the most common abelian group is the integers under addition, and there's something nice about abelian groups (we understand them completely) we often make the leap of writing + for * in abelian groups (personally i think this is stupid, but tradition dictates...)

 

It also means that we need to write 0 for the identity and that we write 2a for a+a.

 

So the first question asks you to show that the set of all elements of order 2 plus the identity is a (sub)group.

 

Associativity of the operation comes for free since it is inherited from the larger group. The identity is there too since 0+0=0, and inverses are there as well since every element is its own inverse. Now, at no point have I used the abelian nature of the group. You must show closure (that the composition of two elements of order two has order 2, or is the identity) and that requires the commutativity - it isn't true for non-abelian groups.

 

 

As for the second one. In an abelian group, if x and y have orders n and m resp, then show that the order of x+y (or x*y) is the least common multiple of n and m.

[matt grime]

Let * denote the group operation. Commutative/Abelian means x*y=y*x for all x and y.

 

Or if you like when you see an expression of the form:

 

x*y*z*w*p*q

 

(which is well defined because of transitivity) then you can rearrange the terms to get it into a better form.

 

Because the most common abelian group is the integers under addition, and there's something nice about abelian groups (we understand them completely) we often make the leap of writing + for * in abelian groups (personally i think this is stupid, but tradition dictates...)

 

It also means that we need to write 0 for the identity and that we write 2a for a+a.

 

So the first question asks you to show that the set of all elements of order 2 plus the identity is a (sub)group.

 

Associativity of the operation comes for free since it is inherited from the larger group. The identity is there too since 0+0=0, and inverses are there as well since every element is its own inverse. Now, at no point have I used the abelian nature of the group. You must show closure (that the composition of two elements of order two has order 2, or is the identity) and that requires the commutativity - it isn't true for non-abelian groups.

 

 

As for the second one. In an abelian group, if x and y have orders n and m resp, then show that the order of x+y (or x*y) is the least common multiple of n and m.