This is problem from I.N Herstein Page 35 Q3 .How should i start doing this ?Hints ? Thanks
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Hint: Consider the left inverse of $a$ and the right inverse of $b$ – PM 2Ring Mar 21 '15 at 09:48
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You have given $abab=aabb$. Try applying some inverses. – Mankind Mar 21 '15 at 09:49
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2@HowDoIMath abab=aabb ,now post multiplying by $b^{-1}$ , i get $aba=aabbb^{-1} $,so it becomes $aba=aab $,now pre multipl by $a^{-1}$ , i get$ ba=ab $...is that right ? – Taylor Ted Mar 21 '15 at 09:52
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I assume that the question said that in G $(ab)^2 = a^2b^2$ for all $a$ and $b$. – PM 2Ring Mar 21 '15 at 09:59
3 Answers
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In groups, you can "cancel", as long as you do it from the same side.
Explicitly:
$ac = bc \implies a = b$, and $ab = ac \implies b = c$
Proof:
$ac = bc \implies (ac)c^{-1} = (bc)c^{-1} \implies a(cc^{-1})= b(cc^{-1}) \implies ae = be \implies a = b$
$ac = bc \implies a^{-1}(ac) = a^{-1}(bc) \implies (a^{-1}a)b = (a^{-1}a)c \implies eb = ec \implies b = c$.
Now you are given that $(ab)^2 = a^2b^2$, that is:
$a(ab)b = a(ab)b$. Use the above.
David Wheeler
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i did as abab=aabb ,now post multiplying by $b^{-1}$ , i get $aba=aabbb^{-1} $,so it becomes $aba=aab $,now pre multipl by $a^{-1}$ , i get$ ba=ab $...is that right ? – Taylor Ted Mar 21 '15 at 09:57
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But how is it valid that in my first step after post multiplying by b inverse i bracketed $bb^{-1}$ togethor .This is not associativity – Taylor Ted Mar 21 '15 at 10:01
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$bb^{-1} = e$, so you can just omit any such pair in a string product of elements: $aabbb^{-1} = aabe = aab$. – David Wheeler Mar 21 '15 at 10:02
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This is valid in multiplication only .Isn't it ? We cannot say anything about other binary operations .Is that right ? – Taylor Ted Mar 21 '15 at 10:04
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It's valid for any GROUP operation. Not all binary operations turn a set into a group. Note the NOTATION for abelian groups MAY be written differently, using $0$ instead of $e$, and denoting $ab$ as $a+b$, and $a^{-1}$ as $-a$, and $a^k$ as $ka$. Group operations are often "called" multiplication, but this is just nomenclature-sometimes group operations are "other things" (like composition of functions). – David Wheeler Mar 21 '15 at 10:41
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I am not sure what is the dot operations and if it's the same as group binary operation, but I will assume that it is, anyway it's group so there could be only one operation. $$ (a \cdot b)^2 = a^2 \cdot b^2 $$
$$ (a \cdot b) (a \cdot b) = a^2 \cdot b^2 $$ $$ a \cdot b \cdot a \cdot b = a \cdot a \cdot b \cdot b $$
First multiply this element by $b^{-1}$ from right and then with $a^{-1}$ from left.
$$ b \cdot a = a \cdot b $$
So it's abeliean group, if my assumtions were good.
fho
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