# A group with a change of operation

let $(G,\cdot)$ be a group, and $g\in G$ we define a new operation on $G$, $a*b=agb$, is $G$ with the new operation is a group?

Associativity:

$(a*b)*c=(agb)*c=(agb)gc$ [because $(G,\cdot)$ is a group]

$(agb)gc=ag(bgc)=a*(b*c)$

So associativity is preserved

Identity element:

We need to prove that there is $e$ such that $a*e=e*a=a$

$age=a$ , if $e$ is the identity element in $(G,\cdot)$ we get that $age=ag$ if not then $ge=e$ so $g$ is the identity element but $a*e\neq ae$

Because there is no identity element, the inverse element axiom can not be defined and so $G$ with the new operation is a magma

Are the claims correct?

## 2 Answers

An identity element should be $x$ such that, for every $a\in G$, $a*x=a$ and $x*a=a$.

Since $a*x=a$, the first relation becomes $agx=a$, which implies $gx=e$ and so $x=g^{-1}$.

Now, what's $a*g^{-1}$? What's $g^{-1}*a$?

You see that an identity element does exist.

What about inverses?

For $a\in G$, you need $b$ such that $a*b=g^{-1}$ and $b*a=g^{-1}$.

We need $agb=g^{-1}$, so $b=g^{-1}a^{-1}g^{-1}$: is it a good candidate for being the $*$-inverse to $a$?

Alternatively, consider the map $\alpha\colon x\to g^{-1}x$, which is a bijection $G\to G$. Define $$a*b=\alpha(\alpha^{-1}(a)\alpha^{-1}(b))$$ This is the standard way to transfer an operation using a bijection, so $*$ is a group operation on $G$.

Now $$a*b=\alpha(gagb)=g^{-1}gagb=agb$$ and it's done.

• Seems to be, I can not find an element that does not have an inverse even $g{-1}$ have an inverse
– gbox
Commented Nov 15, 2016 at 20:45
• @gbox Yes, that's it Commented Nov 15, 2016 at 20:46
• So it is a group, also I tought it is not
– gbox
Commented Nov 15, 2016 at 20:47
• @gbox I added also a much better way. ;-) Commented Nov 15, 2016 at 20:54
• Thanks, but it is to advance for me at this stage :)
– gbox
Commented Nov 15, 2016 at 20:59

You're right about the associativity.

You're wrong about the identity element, because the new identity element $e'$ is not the same as the previous one $e$.

If $a*e'=a$, then $age'=a$ so $ge'=e$ so $e'=g^{-1}$.

Reciprocally, with $e'=g^{-1}$ you do have that $(G,*)$ is a group.

• Ehm, you're jumping to conclusions too fast, I'm afraid. Commented Nov 15, 2016 at 18:37