Groups whose order have order two

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In summary, the conversation discusses how to prove that a group, G, where every non-identity element has order two, is commutative. Hints and examples are provided for approaching the proof, such as experimenting with algebraic expressions and trying simpler cases with different numbers of generators. The forum rules are also mentioned, which state that one must show their attempt at a solution. Additionally, the conversation touches on the topics of subgroups of cyclic groups and whether they are always cyclic.
  • #1
halvizo1031
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I need help here: Suppose that G is a group in which every non-identity element has order two. Show that G is commutative.


Also, Consider Zn = {0,1,...,n-1}
a. show that an element k is a generator of Zn if and only if k and n are relatively prime.

b. Is every subgroup of Zn cyclic? If so, give a proof. If not, provide an example.
 
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  • #2
The forum rules state that you must show your attempt at a solution.
 
  • #3
VeeEight said:
The forum rules state that you must show your attempt at a solution.

I would show an attempt if I knew how to start it.
 
  • #4
Here are some hints:
a) If k generates Zn, then k must have order n. But the order of <k> = n / gcd(k,n).
b) Are subgroups of cyclic groups necessarily cyclic?
 
  • #5
VeeEight said:
Here are some hints:
a) If k generates Zn, then k must have order n. But the order of <k> = n / gcd(k,n).
b) Are subgroups of cyclic groups necessarily cyclic?

b) as far as i know, subgroups of cyclic groups are always cyclic. but to be honest, I do not know if we are allowed to assume Zn is cyclic to begin with. It states that Zn = {0,1,...(n-1)}.

a) so because k generates Zn, generating k (x) amount of times will give us all the elements in Zn?
 
  • #6
halvizo1031 said:
I need help here: Suppose that G is a group in which every non-identity element has order two. Show that G is commutative.
I'm going to assume you've already spent a good amount of time experimenting with algebraic expressions to which you can apply x²=1 in creative ways.

If you haven't, you really should have.


So if the full question is too hard for you, then try a simpler problem first.

First, try to prove it in the case where G has zero generators.
Now, try to prove it in the case where G has one generators.
Now, try to prove it in the case where G has two generators.
Figure it out yet? No? Then try three generators...
 

Related to Groups whose order have order two

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 in the set to produce a third element in the set. The operation must also satisfy certain properties, such as closure, associativity, identity, and invertibility.

What is the order of a group?

The order of a group is the number of elements in the group. It is denoted by |G|, where G is the group. For example, if a group has 5 elements, its order is 5.

What does "order two" mean in the context of a group?

"Order two" in the context of a group means that the group has only two elements: the identity element (usually denoted by e) and another element (usually denoted by a). This also means that the group's order is 2.

What are the properties of groups whose order is two?

Groups whose order is two must have the identity element and one other element. The binary operation must also satisfy the properties of closure, associativity, and identity. Additionally, the other element in the group must be its own inverse (i.e. a * a = e).

What are some examples of groups whose order is two?

Some examples of groups whose order is two include the group of integers modulo 2 (Z2), the group of 2x2 matrices with entries in the field of real numbers (GL(2,R)), and the group of reflections in 2D space (D2).

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