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Let $G$ be a group and $u \in G$ be a fixed element. Define the operation $\bullet$ on G as $\forall a,b \in G, a \bullet b=au^{-1}b.$ Prove that $(G,\bullet)$ is a group.

So, I know that in order for it to be a Group, it must have closure, associativity, identity, and inverse.

Closure is obvious.

Associativity: Let $a ,b ,c \in G$,

then $(a \bullet b) \bullet c=(au^{-1}b \bullet c)$

$au^{-1}bu^{-1}c=au^{-1}(bu^{-1}c)=au^{-1}(b \bullet c)=a \bullet (b \bullet c)$

Identity: If $a,b \in G$ and $e$ is the identity in G, then $a \bullet u=au^{-1}u=ae=a$ and $u \bullet a=uu^{-1}a=ea=a$

Inverse: Now, I believe that the above is correct but Im a little unsure about the inverse.

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Arthur and Squirtle's answers are 100% correct, but just in case you want something more conceptual:

This is an example of transport of structure. In detail, if $(G,\cdot)$ is a group, and $f$ is a bijection from $G$ onto a set $X$, then we can define an operation $\bullet$ on $X$ by setting $$ f(g) \bullet f(h) = f(g \cdot h) $$ for all $g,h \in G$. Note that every element of $X$ has the form $f(g)$ for some $g \in G$, so this operation is well defined on all of $X$.

All of the group properties for $(X,\bullet)$ now follow straight from the properties for $(G,\cdot)$. For instance, $(X,\bullet)$ is associative because $$ ( f(g) \bullet f(h) ) \bullet f(k) = f(g \cdot h) \bullet f(k)\\ = f((g \cdot h) \cdot k)\\ = f(g \cdot (h \cdot k))\\ = f(g) \bullet f(h \cdot k)\\ = f(g) \bullet (f(h) \bullet f(k)).\\ $$

I'll leave you to prove that the identity of $(X,\bullet)$ is $f(\text{id}_{G})$, and the inverse of $f(g)$ is $f(g^{-1})$.

Your example is then $X = G$ (as sets) and $f(g) = ug$, since $$f(ab) = uab = (ua)u^{-1}(ub) = f(a)\bullet f(b).$$

Now write $g^{-1(\bullet)}$ for the inverse of $g$ with respect to $\bullet$, and assume we've proved that $f(e) = u$ is the identity of $(G,\bullet)$. If $g \in G$, then $g = f(u^{-1}g).$ So the inverse property becomes

$$ g^{-1(\bullet)} = f(u^{-1}g)^{-1(\bullet)} = f((u^{-1}g)^{-1}) = f(g^{-1}u) = ug^{-1}u, $$ as in Arthur and Squirtle's answers.

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    $\begingroup$ Cool. I not know it was this general (or even a named phenomenon). $\endgroup$ – Arthur Jan 28 '15 at 4:49
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As for the inverse, given an $a \in G$, we want to find some $b\in G$ such that $a\bullet b = u$. We have $$ a\bullet b = u\\ au^{-1}b = u\\ ua^{-1}au^{-1}b = ua^{-1}u\\ b = ua^{-1}u $$ So the inverse of $a$ in $(G, \bullet)$ is $ua^{-1}u$ (it remains, of course, to show that it is a left inverse as well).

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    $\begingroup$ Well, if we require that $u$ is the identity, then this is correct; otherwise, my solution works. It wasn't stated in the question of the problem so I assumed the identity was the same in both cases. Though it seems a bit strange that $u$ is just "some" element in the opening line, then later called the identity. $\endgroup$ – Squirtle Jan 27 '15 at 22:54
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    $\begingroup$ @Squirtle I have assumed in my answer that $u$ is the identity of $(G, \bullet)$, since that has already been proven. $\endgroup$ – Arthur Jan 27 '15 at 22:57
  • $\begingroup$ Oh yeah, thanks. My bad $\endgroup$ – Squirtle Jan 27 '15 at 22:59
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    $\begingroup$ @Squirtle You cannot have inverses without first having the identity element, since otherwise the whole idea of an inverse is more or less meaningless. $\endgroup$ – Arthur Jan 27 '15 at 23:00
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    $\begingroup$ There is a general theorem that if $\bullet$ is an associative operation on a set $G$ for which $u$ is a left and right identity, and $b'$ is a left inverse for $b$, then $b'$ is also a right inverse for $b$. In general, any three of the four properties (left identity, right identity, left inverse, right inverse) imply the fourth, so that $\langle G, \bullet\rangle$ is a group, and it is a well-known fact that left inverses in groups are the same as right inverses. $\endgroup$ – MJD Jan 27 '15 at 23:41
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What you have above is correct so far. The answer to your last question is a little funny, but The inverse of $b$ is $ub^{-1}u$ where the inverse here is in the first group structure. Explicitly, $(ub^{-1}u)\bullet b=(ub^{-1}u)u^{-1}b=u$; similarly, $b\bullet (ub^{-1}u)=bu^{-1}ub^{-1}u=u$

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