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If $a^6 = e$ and $ab = ba^2$, show that $a^3 = e$ and $aba = b$

Workings:

Proof:

$aba = (ba^2)a$

$aba = ba^3$

$aba = be$

$aba = b$

I'm not sure how to show that $a^3 = e$ though.

Any help will be appreciated.

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  • $\begingroup$ If you don't know how to show $a^3=e$, your proof should not be using that. The question is very confusingly formulated. $\endgroup$ – Marc van Leeuwen Mar 16 '15 at 13:21
  • $\begingroup$ $a^6 = e$ tells you that the order of $a$ divides $6$. $ab = ba^2$ is better written as $b^{-1}ab = a^2$ telling you that $a$ is conjugated to $a^2$ via $b$, which implies that the order of $a$ cannot be divisible by $2$ (conjugated elements have the same order). $\endgroup$ – j.p. Mar 16 '15 at 13:26
  • $\begingroup$ @j.p. That's a (very nice) answer, not a comment. $\endgroup$ – Jack M Mar 16 '15 at 16:40
  • $\begingroup$ @JackM: Thanks, the question had already (for my taste too) many answers. $\endgroup$ – j.p. Mar 16 '15 at 16:56
  • $\begingroup$ This VERY COMMON FIRST YEAR HOMEWORK question is getting a lot of attention, for the non-maths users: maths.kisogo.com/index.php?title=Group <-- what a group is. But seriously, these questions (and how to solve them) is immediately after the definition of a group and some basic properties in (probably) every book on Abstract Algebra $\endgroup$ – Alec Teal Mar 16 '15 at 22:07
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$ab=ba^2 \Rightarrow b^{-1}ab=a^2 \Rightarrow b^{-1}a^3b=a^6=e \Rightarrow a^3=e.$

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    $\begingroup$ Thanks. As you said, I already used the fact that $a^3 = e$ to get $aba = e$ so I couldn't do what they said. $\endgroup$ – CoolNewFriends Mar 16 '15 at 13:17
  • $\begingroup$ @CoolNewFriends: You can't use the statement you're trying to prove as part of its own proof. $\endgroup$ – El'endia Starman Mar 16 '15 at 13:47
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Another way: $b=ba^6=(ba^2)a^4=aba^4=a^2ba^2=a^3b$, and you can now multiply on the right by $b^{-1}$. The identity $ab=ba^2$ gives you a way to cut the exponent on $a$ in half in the right context.

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    $\begingroup$ Nice one. +1 :) $\endgroup$ – Krish Mar 16 '15 at 13:28
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$ab = ba^2 \implies aba = ba^3=b \implies a^3=e$

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  • $\begingroup$ Unfortunately OP used the fact that $a^3=e$ to prove that $b=aba.$ $\endgroup$ – Krish Mar 16 '15 at 13:10
  • $\begingroup$ but it is proved so can be used. $\endgroup$ – Bhaskar Vashishth Mar 16 '15 at 13:23
  • $\begingroup$ No! It's not proved. That was the question. "How to show $a^3=e$?" OP showed here that using this he could prove the second condition. $\endgroup$ – Krish Mar 16 '15 at 13:26
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From $ab=ba^2$ you get $a^2=b^{-1}ab$. Take the third power of that equation and simplify to get $a^6=b^{-1}a^3b$. The left hand side is given to be $e$, so $e=b^{-1}a^3b$ form which $e=a^3$ follows. The rest you already know.

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