Algebra Struggles Part II
by djmathman, Sep 21, 2016, 11:44 PM
I never solved the problem 
I managed to get a bunch of cases but couldn't figure out how to configure things generally rip
It's problem 2 here.
On a related note, we have our first midterm on Friday and our hope is that we don't die - apparently the average on the first test last year was a 50% (which was then curved upward, but still)
I managed to get a bunch of cases but couldn't figure out how to configure things generally rip
chezbgone2 wrote:
hello i think that you should share the problem :
D
DIt's problem 2 here.
and 
are the elements in question, then 
is the desired element. Indeed, this works in certain cases - namely, when 
,
and 
are disjoint. But it doesn't work in general. The counterexample I came up with after 3 hours was the following: let ![\[D_{60} = \langle r,s:r^{30}=s^2 = 1, sr = r^{-1}s\rangle ,\]](http://latex.artofproblemsolving.com/a/5/9/a59be3053903b8466345873c58b63f73d00fa611.png)
and define 
.
and 
satisfy 
and 
,
.On a related note, we have our first midterm on Friday and our hope is that we don't die - apparently the average on the first test last year was a 50% (which was then curved upward, but still)
This post has been edited 2 times. Last edited by djmathman, Sep 22, 2016, 12:38 AM
for all 
with order 
,
.
.
such that![\[k = q n + r \qquad \text{and} \qquad 0 \leq r < n.\]](http://latex.artofproblemsolving.com/c/5/4/c546c76677293ed62906a084b29f3977bd2bfb01.png)
Then we hvae![\[g^k = (g^n)^q g^r = g^r = e.\]](http://latex.artofproblemsolving.com/f/1/2/f12164a25d27819fe1e764b09f17a254b3427085.png)
Thus, 
,
.
.
.![\[g^k = (g^n)^\ell = e.\]](http://latex.artofproblemsolving.com/3/3/c/33cc7c49648402dfce07c5151af07f2ecc5ce859.png)
with orders 
respectively. Then the order of 
divides 
.
.![\[(g_1 g_2)^{\mathrm{lcm}(n_1, n_2)} = g_1^{n_1a} g_2^{n_2b} = (g_1^{n_1})^a (g_2^{n_2})^b = e.\]](http://latex.artofproblemsolving.com/2/e/7/2e7caa49dfe6fd04ecb410291fdf956d807150bb.png)
By Lemma 1, we have the result.
.
.
by Lemma 2. We have![\[a^{nk} = (ab)^{nk} = 1.\]](http://latex.artofproblemsolving.com/d/d/7/dd7db0c1d04cd3fe1b9258fef4b6336bec55d562.png)
Then by Lemma 1, 
.
.
.
.
.
(the prime factorization). Notice that if we can find elements 
that have order 
,
will have order 
from the first part.
or 
has order 
isn't abelian!!!
not 
right?
sending 
to 
.
,
with 
.![$[m, n]$](http://latex.artofproblemsolving.com/b/2/7/b27054efbbe50ae3a72461eef6b45b8aca405e90.png)
in 
so 
April 2025
November 2024