MHB Show that there are y,z such that y,z commute and their order is m and n

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Commute
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

I got stuck at the following exercise:

If $x \in G$ has order $mn$ with $ (m,n)=1 $, show that there are $y,z$ with $ x=yz $ such that $y$,$z$ commute and they have order $m$ and $n$ respectively.

Could you give me some hints?? (Wondering)
 
Physics news on Phys.org
mathmari said:
Hey! :o

I got stuck at the following exercise:

If $x \in G$ has order $mn$ with $ (m,n)=1 $, show that there are $y,z$ with $ x=yz $ such that $y$,$z$ commute and they have order $m$ and $n$ respectively.

Could you give me some hints?? (Wondering)
Hint: think about powers of $x$.
 
Another hint: Since $(m,n) = 1$, there are integers $s$ and $t$ such that $1 = sm + tn$.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 26 ·
Replies
26
Views
688
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
10
Views
2K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K