Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite
1

The order of the smallest possible non trivial group containing elements $x$ and $y$ such that $x^7 = y^2 = e $ and $ yx = x^4 y$ is

  1. 1
  2. 2
  3. 7
  4. 14

I am stuck on this problem. Can anyone help me please?

share|improve this question
1  
Since you are new, I want to give you some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. – Zev Chonoles♦ Jan 2 at 6:32
Are these questions from Subject GRE? – Ram Jan 2 at 6:33
2  
Should we assume both $x$ and $y$ are nontrivial? – anon Jan 2 at 6:36
GRE means,????? – Prasanta Jan 2 at 6:36
2  
I'm with @anon on this. Any group contains such elements $x$ and $y$, because we can always take $x=y=1$. Of course, in some groups there can be other suitable pairs $(x,y)$, but this is irrelevant if the question is formulated as it is. So some kind of an additional non-triviality assumption is required to make the question at least a bit interesting. – Dan Shved Jan 2 at 7:01
show 5 more comments

3 Answers

up vote 2 down vote accepted

Hint:

  1. Lagrange's theorem: For any finite group $G$, the order of every subgroup $H$ of $G$ divides the order of $G$.

  2. $yx = x^4y$; $y(yx) = x$ do the manipulations using $yx = x^4y$ again and again and you will get $x = x^m$ for some $m \le 7 $ use that fact

share|improve this answer
assume $y \neq 0$ $yx = x^4y$ left multiply both sides by $y$ $ x = y x^4y$ $\Rightarrow $ $x = x^4yx^3y$ $\Rightarrow$ $ x = x^4x^4yx^2y$ finally you will get some relation between powers of $x$, you can use that. – Ram Jan 2 at 6:51
1  
@ Ram, I get $x=x^{16}$,so,$x=x^2$,so $x=e$....which is wrong?? – Prasanta Jan 2 at 6:59
1  
If x = e, plug it back into the relations you are given. Can you find out anything new about y? – Billy Jan 2 at 7:07
1  
It would be helpful for me to improve my answers if reason for down vote is provided. I gave this particular method as answer since these type of questions appear very frequently in Indian exams and I felt it would be nice if I give a general procedure. – Ram Jan 2 at 7:15
2  
@Ram I downvoted because I feel that this is not the best approach to the question. There's no need to prove that $x=e$ when we can simply set it to equal $e$. Although I do agree that proving it can also improve one's understanding of the problem. – Dan Shved Jan 2 at 7:21
show 10 more comments
up vote 0 down vote

First solution - $x=y=e$ satisfies the relations. The smallest non-trivial group has order 2, and the relations can be satisfied within that group.

Now suppose we want $x$ and $y$ distinct (not stated in the question).

If $x$ and $y$ are both non-trivial (i.e. $\neq e$) then the first relation shows that the group must contain non-trivial elements of orders 2 and 7, and then Lagrange means that the order of such a group must be divisible by 14. [note we have not used the second relation or shown it is compatible with this conclusion]

So to get a non-trivial group of order less than 14, one of $x$ or $y$ must be the identity. If we set $y=e$ we see that $x^7=x^3=1$ so that $x=e$, which is not what we want. If we set $x=e$ then $y^2=e$ and this can be done in a group of order 2.

share|improve this answer
up vote 3 down vote

Up to isomorphism, there is only one group of order 1, one group of order 2, one group of order 7, and two groups of order 14. Figure out (or look up) what those groups are. Which ones are non-trivial? Look for suitable elements $x$ and $y$ in those groups. You may want to start from the smallest group and work your way up.

share|improve this answer
Nice answer, but I doubt that might be little advanced for him right now. – Ram Jan 2 at 6:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.