2
$\begingroup$

Let $G$ be a group and $a_i \in G$ for all $ n \in N$.

  1. Prove that $(a_1a_2a_3)^{-1} = a_3^{-1}a_2^{-1}a_1^{-1}$

My Solution: $(a_1a_2a_3)^{-1}$ = $(a_1a_2)^{-1}(a_3)^{-1}$ = $a_1^{-1}a_2^{-1}a_3^{-1}$

(Now as only Abelian groups are commutative then how can the order change in the answer. Either that or I'm missing something.)

$\endgroup$
2
  • $\begingroup$ Distributivity properties only necessary hold in abelian groups, too (ie. $(ab)^c=a^cb^c$) $\endgroup$ Dec 29, 2014 at 21:11
  • 1
    $\begingroup$ No, it doesn't work that way, there is no rule justifying the operations that you do... You have to show that $a_3^{-1}a_2^{-1}a_1^{-1}$ is the inverse of $a_1a_2a_3$. This means you must show that their product equals the identity element... $\endgroup$
    – Myself
    Dec 29, 2014 at 21:16

2 Answers 2

3
$\begingroup$

$a_1 a_2 a_3 a_3^{-1} a_2^{-1}a_1^{-1}=1$ shows the result by definition.

$\endgroup$
3
$\begingroup$

I solved my question. Missed such a simple thing.

$(ab)^{-1} = b^{-1}a^{-1}$.

Through this one is able to solve the question.

$(abc)^{-1} = [(ab)(c)]^{-1} = (c)^{-1}(ab)^{-1} = c^{-1}b^{-1}a^{-1}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .