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May 16th, 2017, 12:12 PM  #1 
Newbie Joined: May 2017 From: Neverland Posts: 7 Thanks: 0  [Set theory]  countering(oppositing)
K is a Group in relation to *, and a,b,c are different parameters that belong to K. and given this equation a*b = b*c. I need to prove that if a counters itself, then c does too. I can't really express how it works... may I have some guidance here please? Thanks. Sorry for bad English. Last edited by skipjack; May 17th, 2017 at 05:59 AM. 
May 16th, 2017, 12:21 PM  #2  
Senior Member Joined: Aug 2012 Posts: 1,765 Thanks: 480  Quote:
$(K, *)$ is a group, $a, b, c \in K$. If $ab = bc$ then if $a = a^{1}$ then $b = b^{1}$. [Ignoring the * entirely. Don't need it]. Last edited by skipjack; May 17th, 2017 at 06:00 AM.  
May 16th, 2017, 12:27 PM  #3  
Newbie Joined: May 2017 From: Neverland Posts: 7 Thanks: 0  Quote:
And yes I meant selfinverse, but didn't get it how is b = b^{1} ? I've only started studying set theory 3 weeks ago, so I'm not that advanced yet... Last edited by skipjack; May 17th, 2017 at 06:01 AM.  
May 16th, 2017, 01:19 PM  #4  
Senior Member Joined: Aug 2012 Posts: 1,765 Thanks: 480  Quote:
We have $ab = bc$. Ok, what can we do? I just tried the most mindless thing, I multiplied both sides on the left by $a$ to get (1) $aab = abc$. Under the assumption that $a = a^{1}$ we have $a^2 = aa =1$ so (2) $aab = abc \implies b = abc$. Since $ab = bc$ (given) we can write $(ab)c = (bc)c = bc^2$ using associativity (referring back to (2), so now we have $b = bc^2$. (Combining (1) and (2)) And then hit it hard with $b^{1}$ on the left to get $b^{1}b = b^{1}bc^2 \implies 1 = c^2 \implies c = c^{1}$. I freely used associativity a few times without mentioning it but you should notice where I did it. This is the general pattern for a lot of these kinds of problems. You just start multiplying both sides by this or that and shuffle things around till the result falls out. After a while you get better at it. Don't worry about insight, just learn to crank the symbols. Insight comes much later when you need to use groups for something specific. Do be aware that group theory is everywhere in math. There is light at the end of the abstract algebra tunnel Last edited by skipjack; May 17th, 2017 at 06:03 AM.  
May 16th, 2017, 01:33 PM  #5  
Newbie Joined: May 2017 From: Neverland Posts: 7 Thanks: 0  Quote:
I've got the picture sun bright, I'll treasure your advice! Many thanks! Last edited by skipjack; May 17th, 2017 at 06:14 AM.  

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