Order of Elements in a Group | Abstract Algebra

We introduce the order of group elements in this Abstract Algebra lessons. We'll see the definition of the order of an element in a group, several examples of finding the order of an element in a group, and we will introduce two basic but important results concerning distinct powers of elements with finite order and elements with infinite order. #abstractalgebra #grouptheory
Permutation Groups: • Infinite Order Element...
Finding the Order of Group Elements: • Finding the Order of G...
Finite Powers of an Element are Distinct: • Proof: Finite Order El...
Infinite Order Elements have Distinct Powers: • Infinite Order Element...
Abstract Algebra Course: • Abstract Algebra
Abstract Algebra Exercises: • Abstract Algebra Exerc...
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[@WrathofMath]

I previously had a video on this, but I used a needlessly complicated definition which made the lesson a few minutes longer than it needed to be. So I redid it.

[@adhithugs2.094]

Nicely Explained✨

[@keldonchase4492]

Thanks for the videos; they are immensely helpful!!! Question about the example presented at 4:40: The way I look at it is: Cycle 1: Start with 1: 1 --> 6 (1) 6 --> 4 (2) 4 -> 2 (3) 2 -> 1 (4) I started with 1 and ended with 1 so to get back to 1, I had to make 4 “jumps” so 4 is a possible answer for our order. Cycle 2: Skip all the numbers covered in Cycle 1 so I start with 3. 3 --> 3 (0) Since 3 maps to itself, I do not consider a “jump”. 0 would be a possible answer but since it is non-positive, I can eliminate it as a possibility. Cycle 3: 5 --> 5 (0) Again, not a “jump” because 5 maps to itself. Again, 0 can be eliminated as a possibility since it is non-positive. So that is why the order is 4. Is this way of thinking correct or is there something important I’m missing? Thank you!!

[@faysal_ahamed2701]

Thanks sir , I try to improve math skills by your video

[@SaiDharahasReddyIndrakanti]

6:47 will the order be 3? because you are starting with 1->6 but i think we included it in while counting

[@blue-cj6bw]

thanksssss😭💘💘💘💘

[@Dravignor]

If e is the identity element of G, and e is also a power of some a ∈ G, then can we say ord(e) = 0, or is it ord(e) = 1 because the order of an element is strictly for non-zero positive integers? Edit: Nevermind I watched the next video

[@MrCoreyTexas]

I was wondering why not 0 for the order, and the reason it is not 0 is because every element to the 0th power is the identity element, so it has to be a positive integer to be non trivial

[@mgyodzs1]

Good explanation! Keep going, You do good job.

[@WrathofMath]

Thanks, will do!

[@jialinding9636]

The frog is the icing on the cake.

[@WrathofMath]

It always is!

[@iraqi-ff9690]

Thank u so much 🖤

[@WrathofMath]

You're welcome!

[@summerhunt77]

Thank you.

[@WrathofMath]

Glad to help!

[@HorenKriz]

So helpful thanks

[@WrathofMath]

Thanks for watching!

[@punditgi]

You can never go wrong with Wrath of Math!

[@WrathofMath]

Such is the order of things!

[@NathalieBertrandCortez]

very interesting

[@WrathofMath]

I think so too!

[@InoceramusGigas]

Hi W.O.M Would love some videos on ramsey theory! Could be a great fit in your graph theory playlist, or just within a general combinatorics theme. There is definitely room for a more intuitive explanation on KZbin. Love the Abstract algebra vids.

[@naruhitoabiku9451]

i love you

[@tjstarr2960]

The abuse of notation really doesn't help in this video. I was wondering why you suddenly switched from saying the order is "n" where "a * n = e", to saying "a^n = e". Those are two completely different operations, multiplication vs. exponentiation! But, then I realized that Group Theory applies generally to any set with an associated binary operation operation. So, in the beginning, when you said a * n = e, you were talking about Groups with the Addition operation +. So, using the proper notation for Groups, for the first example, where you wrote 3 ∈ ℤ6, you really meant it belonged to the additive group 3 ∈ (ℤ6, +). In this case, we are asking how many copies of 3 do we need to get to the identity, but because the operation is addition, we are asking how many times do we have to ADD 3 together to get the identity, which is the same as asking what do we have to multiply by 3 to get the identity. But, in the case you were saying the order of the element "a" was "n" where "a^n = e", you were talking about Groups with the multiplication operation, written as (G, *) . So, you are still asking the question "How many copies of 3 do I need to combine to get the identity", but since the operation in this case is multiplication, we are asking how many times do we have to MULTIPLY 3 together to = e. The number of times we multiply a number is the same as taking the exponent to that power. Please let me know if I am right about this, I am still learning this subject myself.

[@smbushi]

Isn't a group technically a set of ordered pairs? So it is wrong to say that 3 is an element of (Z6,+). 3 is an element of Z6 which is a group under addition. Also I believe * is reserved for arbitrary operations while multiplication preserves its dot symbol. Also when he was discussing examples in order he said he wrote them in multiplicative notation; not necessarily referring to the operations multiplication or addition specifically. This I believe just means a*b*c. Regarding powers (a^n), I believe that they are raised to n under the operation of the group. I.e if the operation was o then a^2=a o a. But I'm only in 9th grade so I might as well be even more wrong than you are.

[@AIstudentsoon]

am i the only one who couldn't understand this 😭😭? I'm feeling very dumb, i watched it so many times and still can't get it well idk why🥹!! ( I'm a math major)

[@blue-cj6bw]

thanksssss😭💘💘💘💘

[@WrathofMath]

Glad to help, thanks for watching!