Intereting Posts

Given a pairwise disjoint collection, $\limsup A_n = \emptyset$?!
Series with fractional part of $nx$
Proof of $\sin nx=2^{n-1}\prod_{k=0}^{n-1} \sin\left( x + \frac{k\pi}{n} \right)$
Why square matrix with zero determinant have non trivial solution
What's an example of a number that is neither rational nor irrational?
When not to treat dy/dx as a fraction in single-variable calculus?
How to find $\omega^7$ and $\omega^6$ from $\omega^5+1=0$
$\kappa$-complete, $\lambda$-saturated ideal equivalence
Structure sheaf consists of noetherian rings
Getting Students to Not Fear Confusion
How to find the vector equation of a plane given the scalar equation?
Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$
How $a^{\log_b x} = x^{\log_b a}$?
$\dim C(AB)=\dim C(B)-\dim(\operatorname{Null}(A)\cap C(B))$
Yoneda's lemma and $K$-theory.

Let $(G,*)$ be a group with identity $e$ , let $a,b∈G$ such that $a*b^3*a^{-1}=b^2$ and $b^{-1}*a^2*b=a^3$ , then how do we prove that $a=b=e$ ?

- Homomorphisms and exact sequences
- Transcendental number
- Determine the center of the dihedral group of order 12
- Showing the polynomials form a Gröbner basis
- Prove that any group $G$ of order $p^2$ is abelian, where $p$ is a prime number
- Yoneda-Lemma as generalization of Cayley`s theorem?
- Is it true that $(R\times S)\cong R\times S$?
- centralizer of transvection
- Exact sequences of $SU(N)$ and $SO(N)$
- $R/\mathfrak p$ not always a UFD

Given $$ab^3a^{-1}=b^2$$ thus we get $$ab^6a^{-1}=b^4\;,\;\;ab^9a^{-1}=b^6$$

Replaceing $b^6$ in $ab^6a^{-1}=b^4$, we get $a^2b^9(a^{-1})^2=b^4$. Thus $a^2b^{18}(a^{-1})^2=b^8$. Again $ab^3a^{-1}=b^2$ gives $ab^{27}a^{-1}=b^{18}$. Replacing $b^{18}$, we get $a^3b^{27}(a^{-1})^3=b^8$. But $b^{-1}a^2b=a^3$. Thus $(b^{-1}a^2b)b^{27}(b^{-1}a^2b)^{-1}=b^8$. This gives $a^2b^{27}(a^2)^{-1}=b^8$. Also $a^2b^9(a^{-1})^2=b^4$ and $a^2b^{18}(a^{-1})^2=b^8$ give $a^2b^{27}(a^2)^{-1}=b^{12}$. Thus $b^8=b^{12}, i.e. b^{4}=e$. Thus $ab^6a^{-1}=e$, and thus $b^6=e$. so, $b^2=e$ and similarly it can be shown that $b^3=e$. So, $b=e$. Now showing $a=e$ is easy.

Here is a visualization of Anupam’s proof. I think that this method is also applicable for other groups.

A group is just a category with one object in which each morphism is invertible. In categories we can draw commutative diagrams. The given relations can be drawn as follows:

Here $a$ represents the red vertical arrow and $b$ the green horizontal arrow. Now we reproduce the first diagram $4$ times horizontaly and then $3$ times vertically, in order to align three red arrows:

The outer rectangle can be twisted with the second relation, which yields:

But the large diagram above also contains this, but with $12$ instead of $8$ green arrows below. Thus, $4$ green arrows cancel. From parts of the large diagram we see that then $6$ green arrows cancel, hence also $2$ cancel. But then $3$ cancel, and thus the green arrow cancels. The second relation tells us that the red arrow also cancels. QED

Edit: I seem to have reinvented van Kampen diagrams.

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