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 May 16th, 2017, 12:12 PM #1 Newbie   Joined: May 2017 From: Neverland Posts: 8 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
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 Originally Posted by Adm 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.
No prob. By "counters itself" I assume you mean self-inverse. So the question is

$(K, *)$ 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
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 Originally Posted by Maschke No prob. By "counters itself" I assume you mean self-inverse. So the question is $(K, *)$ 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].
Thank you for replying sir!
And yes I meant self-inverse,
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
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 Originally Posted by Adm Thank you for replying sir! And yes I meant self-inverse, 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...
Most of the time you just bang these out whether or not there's insight. Groups are very simple, one operation. There's so little you can do, that you might as well just try things.

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
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 Originally Posted by Maschke Most of the time you just bang these out whether or not there's insight. Groups are very simple, one operation. There's so little you can do, that you might as well just try things. 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 $aab = abc$. Under the assumption that $a = a^{-1}$ we have $a^2 = aa =1$ so $aab = abc \implies b = abc$. Since $ab = bc$ (given) we can write $(ab)c = (bc)c = bc^2$ using associativity, so now we have $b = bc^2$. 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 note those. 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. Don't worry about insight, just learn to crank the symbols, and don't worry about insight. That comes much later when you need to use groups for something specific.
My god I've spent 30 mins trying to figure out something and didn't come to just trying adding same elements to each side for some reason and given up...
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. Tags counteringoppositing, set, theory Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rohantomar10 Number Theory 6 October 27th, 2017 05:21 PM flextera New Users 0 July 30th, 2014 12:12 PM gaussrelatz Algebra 3 August 8th, 2012 09:35 AM proglote Number Theory 3 October 30th, 2011 04:20 PM butabi Abstract Algebra 8 September 3rd, 2011 01:52 PM

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