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Ok so I am looking at a proof to all the groups of order 8, I've attached an image which basically is the start of the proof. In particular the part highlighted with yellow is causing me problem, I can see why it works (by subbing it into previous equations) but without being told a is of this form, how do you get an insight to write a in this form?

Thanks

Raul
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    $b^2 \in \langle a \rangle$, so $b^2$ commutes with $a$ – zcn Jan 28 '14 at 18:51
  • @user115654 ah right now its clear. :) thanks – Raul Jan 28 '14 at 18:56
  • There is an error in the proof. In the $C_2\times\ C_2\times C_2$ case, "$a\in G\setminus{e}$, $b\in G\setminus {a}$ and $c\in G\setminus{b}$" should be replaced with "$a\in G\setminus{e}$, $b\in G\setminus {e, a}$ and $c\in G\setminus{e, a, b}$". – Hagen von Eitzen Jan 28 '14 at 19:01
  • @HagenvonEitzen are you denoting {a} as the set generated by a? yes the above is wrong as I just noticed, but the proof I have in hand is actually edited and correct, thanks for looking at the proof :) thanks – Raul Jan 28 '14 at 19:04
  • which book is this? – AG. Feb 01 '14 at 12:34

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