Have a proof re. cyclic groups, need a little explaining

In summary, the author argues that if a group has finite order, then its subgroups have finite order as well. He provides a proof.
  • #1
bennyska
112
0

Homework Statement


Let a,b be elements of a group G. show that if ab has finite order, then ba has finite order.


Homework Equations





The Attempt at a Solution


provided proof:
Let n be the order of ab so that (ab)n = e. Multiplying this equation on the left by b and on the right by a, we find that (ba)n+1 = bea = (ba)e. Cancellation of the first factor ba from both sides shows that (ba)n = e, so the order of ba is < n. If the order of ba were less than n, a symmetric argument would show that the order of ab is less than n, contrary to our choice of n. Thus ba has order n also.

okay. so, this is from a chapter on cyclic groups, so I'm assuming that it has to do with, duh, cyclic groups. i know cyclic groups are abelian, and if that were to be assumed from the beginning, i believe this problem would be relatively easy. although, actually, it probably wouldn't even be a problem, since it would be true from the beginning (since we could just say ab=ba, and abn=ban

so:
Let n be the order of ab so that (ab)n = e.
got that, that makes sense, i believe that's the definition of order of an element.
WAIT. here's the crux, something i (maybe) just realized as i typed this sentence. since we define the order of ab such that abn=e, does that imply that G is cyclic, and thus abelian? that would make things easier, if I'm right. so far, a quick internet search has failed me on an answer.
 
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  • #2
If the group is abelian then ab=ba and there is really nothing to prove. The proof you gave holds regardless of whether the group is abelian or not.
 
  • #3
okay, i was just confused as how b(ab)na could turn into (ba)n+1 without it being abelian. i see how it's from letting (ab)n = e. thanks.
 
  • #4
bennyska said:
okay, i was just confused as how b(ab)na could turn into (ba)n+1 without it being abelian. i see how it's from letting (ab)n = e. thanks.

It's actually just from regrouping the terms.
 
  • #5
yeah, actually, i still don't get it. how can you regroup the terms if it's not commutative? or at least it's not assumed to be commutative.
we have (ab)(ab)...(ab) n times. so b(ab)na = b(ab)(ab)...(ab)a...
Oh! got it! thanks!
 

Related to Have a proof re. cyclic groups, need a little explaining

1. What is a cyclic group?

A cyclic group is a group that is generated by a single element. This means that all the elements in the group can be obtained by repeatedly applying the group operation to the generator element. In simpler terms, a cyclic group is a group that forms a cycle when its elements are multiplied together.

2. How do you prove that a group is cyclic?

To prove that a group is cyclic, you need to show that there exists an element in the group that can generate all other elements through repeated application of the group operation. This can be done by finding an element that, when raised to different powers, produces all the elements in the group.

3. What is the order of a cyclic group?

The order of a cyclic group is the number of elements in the group. This is equal to the number of times the generator element needs to be multiplied with itself to produce all the elements in the group. The order of a cyclic group is always finite.

4. Can a cyclic group be infinite?

Yes, a cyclic group can be infinite. This happens when the generator element has infinite order, meaning that it can be multiplied with itself an infinite number of times to produce all the elements in the group. An example of an infinite cyclic group is the group of all integers under addition.

5. How are cyclic groups used in mathematics?

Cyclic groups are used in various areas of mathematics, including algebra and number theory. They are particularly useful in studying symmetry and patterns, and in solving equations in abstract algebra. They also have applications in cryptography, as they can be used to generate large, random numbers.

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