Your argument works for Z/8, but there are 5 groups of order 8 (uti): Z/8, Z/2 x Z/4, Z/2 x Z/2 x Z/2, D8, and Q. Z/2 x Z/2 x Z/2 is of interest here. The group is also a vector space over Z/2, and the automorphisms correspond to matrix group SL(3, Z/2). An element of order 7 is given by [0 0 1 / 1 0 1 / 0 1 0]. Hope that helps!
Much better. It doesn't say action by automorphisms, in which case T. Fix any element, and permute the other seven elements nontrivially. This generates a nontrivial Z/7 action. Note that we are only using G as a set, not a group. If by automorphisms, the statement is a little vague about whether they mean all or at least one.
I would like contribute with an (almost trivial) proposition about what happens to the order of the elements when you take the m-th power. At 2:55 you say "if I take an element of order 4 and square it, I get an element of order 2". This is true but, in general, squaring elements (or taking any power by a positive integer) doesn't necessarily decrease the order. So I tried to investigate this fact and I easily found out that only the divisors of the order are the powers that I was interested in, because we can factor the order, always getting the identity as result. A good instrument for working with divisors is the greatest common divisor, so after some trial and error I came up with the following generality: PROPOSITION: if x in G is an element of order n, then x raised to the m-th power has order n/d where d := gcd(n,m). Proof: If x^m has order k, we can write x^(k*m) = (x^k)^m = e = x^n and this means that n has to divide k*m (because n is the order of x, see below). Because of the definition of the gcd, there are two integers a and b such that n = d*a, m = d*b and gcd(a,b) = 1. So, we can affirm that d*a divides k*d*b and this implies that a divides k*b. Recall that gcd(a,b) = 1.This means that a has to divide k. We can also write (x^m)^a = x^(m*a)= x^(d*b*a) = x^(n*b) = (x^n)^b = e^b = e = (x^m)^k and this implies that k divides a (k is the order of x^m). So a divides k and vice versa, and we can conclude that k = a = n/d. To be precise, this proposition requires a lemma: if n is the order of x and x^t = e, then n divides t. In fact, we can write t in the following way (by Euclidean algorithm): t = q*n + r where q,r are integers and 0
@MathDoctorBob Hi! I have a question I need help with: RE: generators and relations; how do we know when to recognise certain elements as sufficient for a generating set, and likewise, how do we know what relations are sufficient to guarantee the group? With this, at 6:05 you list the three relations that make D_4 (D_2*4); how do we know that that these are the only relations we need to account for? I'll keep saying this... these videos are really helpful! Thank you so much!!
Your welcome - glad they are helping! In general, group presentations can be difficult to work with - one bumps up against the word problem - whether two products of generators represent the same element or not. The field of study is combinatorial group theory. Dihedral groups are fairly special. Two generators, say c and r, and their interchange (rc=cr^-1) is as nice as could be. The key is the interchange, which allows us to convert any word in c and r as c^i r^j. In this way, there is no doubt whether two elements, written in any form as a product of generators, are the same or not.
You can alternatively prove 2:35 by using the result that the product of a finite abelian group must equal the unique element of order 2 if it exists. Taking the product of the group gives you: e x x^2 x^3 y yx yx^2 yx^3 = e (x x^3) (x x^3) (x^2 x^2) y^4 = y^4 From the result, we have y^4 = x^2 but the order of y is 4 by assumption so y^4 = e. This means that x^2 = e, which is a contradiction.
I should have phrased the question as it is: T/F: Integers modulo 7 can act nontrivially on a group G of order 8. I guess I am confused as to what the question is asking...Is the question asking if there is a particular group G or is it asking if it is true for all groups G of order 8. Sorry to bother you with this, English is not my first language.
Your argument works for Z/8, but there are 5 groups of order 8 (uti): Z/8, Z/2 x Z/4, Z/2 x Z/2 x Z/2, D8, and Q.
Z/2 x Z/2 x Z/2 is of interest here. The group is also a vector space over Z/2, and the automorphisms correspond to matrix group SL(3, Z/2). An element of order 7 is given by [0 0 1 / 1 0 1 / 0 1 0].
Hope that helps!
Much better. It doesn't say action by automorphisms, in which case T. Fix any element, and permute the other seven elements nontrivially. This generates a nontrivial Z/7 action. Note that we are only using G as a set, not a group.
If by automorphisms, the statement is a little vague about whether they mean all or at least one.
I would like contribute with an (almost trivial) proposition about what happens to the order of the elements when you take the m-th power.
At 2:55 you say "if I take an element of order 4 and square it, I get an element of order 2". This is true but, in general, squaring elements (or taking any power by a positive integer) doesn't necessarily decrease the order. So I tried to investigate this fact and I easily found out that only the divisors of the order are the powers that I was interested in, because we can factor the order, always getting the identity as result. A good instrument for working with divisors is the greatest common divisor, so after some trial and error I came up with the following generality:
PROPOSITION: if x in G is an element of order n, then x raised to the m-th power has order n/d where d := gcd(n,m).
Proof: If x^m has order k, we can write x^(k*m) = (x^k)^m = e = x^n and this means that n has to divide k*m (because n is the order of x, see below). Because of the definition of the gcd, there are two integers a and b such that n = d*a, m = d*b and gcd(a,b) = 1. So, we can affirm that d*a divides k*d*b and this implies that a divides k*b. Recall that gcd(a,b) = 1.This means that a has to divide k.
We can also write (x^m)^a = x^(m*a)= x^(d*b*a) = x^(n*b) = (x^n)^b = e^b = e = (x^m)^k and this implies that k divides a (k is the order of x^m).
So a divides k and vice versa, and we can conclude that k = a = n/d.
To be precise, this proposition requires a lemma: if n is the order of x and x^t = e, then n divides t. In fact, we can write t in the following way (by Euclidean algorithm): t = q*n + r where q,r are integers and 0
@MathDoctorBob Hi! I have a question I need help with:
RE: generators and relations; how do we know when to recognise certain elements as sufficient for a generating set, and likewise, how do we know what relations are sufficient to guarantee the group? With this, at 6:05 you list the three relations that make D_4 (D_2*4); how do we know that that these are the only relations we need to account for?
I'll keep saying this... these videos are really helpful! Thank you so much!!
Your welcome - glad they are helping! In general, group presentations can be difficult to work with - one bumps up against the word problem - whether two products of generators represent the same element or not. The field of study is combinatorial group theory.
Dihedral groups are fairly special. Two generators, say c and r, and their interchange (rc=cr^-1) is as nice as could be. The key is the interchange, which allows us to convert any word in c and r as c^i r^j. In this way, there is no doubt whether two elements, written in any form as a product of generators, are the same or not.
Cool. Thanks for the reply mate.
Thanks!
You can alternatively prove 2:35 by using the result that the product of a finite abelian group must equal the unique element of order 2 if it exists. Taking the product of the group gives you:
e x x^2 x^3 y yx yx^2 yx^3
= e (x x^3) (x x^3) (x^2 x^2) y^4
= y^4
From the result, we have y^4 = x^2 but the order of y is 4 by assumption so y^4 = e. This means that x^2 = e, which is a contradiction.
I should have phrased the question as it is: T/F: Integers modulo 7 can act nontrivially on a group G of order 8. I guess I am confused as to what the question is asking...Is the question asking if there is a particular group G or is it asking if it is true for all groups G of order 8. Sorry to bother you with this, English is not my first language.
Can the integers modulo 7 act nontrivially on a group G of order 8? I think not since Aut(G) is isomorphic to Z_2 x Z_2. Am I correct? Thanks.
sir ...i have a question there are only two groups of order six one is cyclic and the other is somorphic to s subscprit 3???
You're welcome!
why finite group's all element are not distinct?sir plz help me
Need to clarify. What do you mean by distinct?
distinct means disjoint
thanks Dr Bob
Thanks!