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In Non-abelian group of order $6$, a nonabelian group of order 6 was constructed as: $ \{1,a,b,ab,ba,aba\} $ where $1$ is the identity and $a$ and $b$ are their own inverses. Since the group is nonabelian, $ab≠ba$. If we define $aba=bab$, then the group is indeed closed under multiplication:

* 1 a b ab ba aba
1 1 a b ab ba aba
a a 1 ab b aba ba
b b ba 1 bab=aba a baba=ab
ab ab aba a abab=ba 1 ababa=b
ba ba b bab=aba 1 baba=ab a
aba aba ab abab=ba a ababa=b 1

My question is: why do we have to have $aba=bab$? In the linked question, the answer says $aba=bab$ "by symmetry", but I don't understand what that means. Are we simply defining $aba$ to equal $bab$, or is there something more interesting going on here?

Thanks!

plshelpme1234
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1 Answers1

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Clearly $bab$ is neither $1$ nor $a$ (because then $ba=ab^{-1}=ab$) nor $b$ nor $ab$ nor $ba$. Therefore, it's the last element; by the same logic, so is $aba$.

J.G.
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  • Thank you; I am still wondering how we know that bab isn't a new element entirely? Are you assuming that {1,a,b,ab,ba,aba} must be a nonabelian group? – plshelpme1234 Dec 30 '20 at 22:42
  • plshelpme1234 If an order-$6$ group has identity $1$ and $a\ne b$ both differ from $1$ then $ab,,ba$ are distinct again, and $aba$ is in the group by closure, as is $bab$. The point is each is required to the the only unknown element. – J.G. Dec 30 '20 at 22:43