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I've come across another rather simple question, but I'm having trouble with $G$'s defined operation.

Let $(G,\circ)$ be a group and $x_0 \in G$. Consider the operation $*$ defined in $G$ by $a * b = a \circ x_0 \circ b, a,b\in G$.

a) Knowing $*$ is associative, show $(G,*)$ is a group.

Well, so far I've got this:

  1. Since $x_0 \in G$, G is non-empty.
  2. Closed under multiplication: Let $a,b \in G$. $a * b = a \circ x_0 \circ b$. Since all $a, b,$ and $x_0$ are in $G$, then its product is in $G$ as well.
  3. Identity: $a * 1_G = a$ (I don't think I've got this one quite right, because I'm messing with inverses, and that's probably not required)

$a * 1_G = a == a \circ x_0 \circ 1_G = a == a^-1 \circ a \circ x_0^-1 \circ 1_G = a^-1 \circ a == x_0 \circ 1_G = 1_G == x_0 = 1_G$

I'm left with proving the inverse element, which should probably be done before the identity. Or maybe I should re-think my identity strategy. I'm a bit lost here. Any help? Thanks.

EDIT: What I mean is: Clearly, we need the concept of inverses to work out the group's identity element. Thing is, I proved it before proving the very existence of the inverse element. Is this not a problem?

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  • $\begingroup$ Well thank you Norma! $\endgroup$ – AJ44 Jan 21 '16 at 16:29
  • $\begingroup$ In particular, I can't read your definition of the operation $\endgroup$ – Ross Millikan Jan 21 '16 at 16:30
  • $\begingroup$ Sorry, but I can't follow your calculation. I tried to improve your notation, but there are too many thinks, where I don't know what you mean... $\endgroup$ – user302982 Jan 21 '16 at 16:31
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    $\begingroup$ $a \star x_0^{-1} = a \circ x_0 \circ x_0^{-1} = a$ $\endgroup$ – AnalysisStudent0414 Jan 21 '16 at 16:37
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    $\begingroup$ @AJ44 no it is exactly the other way around. You don't need inverses to talk about the indentiy element at all. The idenity element $e$ is defined as the element such that $a*e = a = e*a$. In fact you need the identity element to define what the inverse is! $\endgroup$ – Jorik Jan 21 '16 at 16:49
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The problem you are having with the identity element is that $1_{(G,\circ)}$ is not the identity element in $(G,\ast)$, $x_0^{-1}$(with inverse taken with respect to $(G,\circ)$) is since $$a\ast x_0^{-1}=a\circ x_0\circ x_0^{-1}=a\circ 1_{(G,\circ)}=a$$ and $x_0^{-1}\ast a=a$ follows almost identically.

With this in mind, can you figure out what the inverses are?

Also, you mention that you think you should find inverses before finding the identity. This is ill-advised because inverses are defined in terms of the identity, so without identity, the concept of inverses doesn't mean anything.

Edit: to show that $(G,\ast)$ is closed under $\ast$, we can use the fact that $(G,\circ)$ is closed under $\circ$. Explicitly, for any $a,b\in G$, $$a\ast b=a\circ x_0 \circ b\in G$$ since $\circ$ is a binary operation on $G$.

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  • $\begingroup$ you and @Meitar brought up some interesting points. Let me think for a sec. Thanks! $\endgroup$ – AJ44 Jan 21 '16 at 16:57
  • $\begingroup$ I got the inverses figured out. Thanks a lot! As for the definition of inverses, you're right. I should be more careful next time. $\endgroup$ – AJ44 Jan 21 '16 at 17:08
  • $\begingroup$ No problem. To really get a grasp on the nature of inverses and their relation to identity, it might not hurt to look up an example of a "monoid" that isn't a group. A monoid is a structure that has an associative binary operation with identity, but does not necessarily have inverses. The existence of such structures concretely shows that the idea of an identity element does not rely on the idea of inverses at all. $\endgroup$ – Sean English Jan 21 '16 at 17:19
  • $\begingroup$ Sorry to ping you again @Sean English. But what about the closed under multiplication? I'm not sure whether or not that's enough. I can't help but think that any two elements from G could produce a result out of G. How do I properly prove it? $\endgroup$ – AJ44 Jan 21 '16 at 17:20
  • $\begingroup$ This follows immediately from the fact that $\circ$ is closed under multiplication. $\endgroup$ – Sean English Jan 21 '16 at 17:21
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As you can see, $a *1_G=a\iff x_0 \circ 1_{G_2}=1_{G_1}$ where $G_1=(G,\circ)$ and $G_2=(G,*)$. Hence you get $1_{G_2}={x_0^{-1}}_{G_1}$. That would mean the inverse of $x_0$ under the operation "$\circ$" is the identity element in $G_2$, which is $G$ under $*$ operation.

As for finding the inverse, for a given $a\in G$, we want to find $b$ such that $a*b=1_{G_2}={x_0^{-1}}_{G_1}$. So, $a*b=a \circ x_0 \circ b={x_0^{-1}}_{G_1}$.

Noticing that the last line is simply an expression taking place in $G_1$, under that operation "$\circ$", we can proceed, noting ${y}^{-1}_{G_1}={y^{-1}}$, assuming we are already making steps in $G_1$: $a \circ x_0 \circ b={x_0^{-1}}_{G_1}\Rightarrow x_0 \circ b=a^{-1}{x_0^{-1}}\Rightarrow b={x_0^{-1}}a^{-1}{x_0^{-1}}$, which, getting back to $G_2$, will be denoted: $a^{-1}_{G_2}=b=({x_0^{-1}}_{G_1})(a^{-1}_{G_1})({x_0^{-1}}_{G_1})$. It is very important to notice that $a^{-1}_{G_2}$ and $a^{-1}_{G_1}$ are not necessarily the same.

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As already noted, Identity element in $(G,*)$ is $x_0^{-1}$.

The inverse of an element $a \in (G,*)$ is $x_0^{-1} \circ a^{-1}\circ x_0^{-1}$:

$$ a \ast (x_0^{-1} \circ a^{-1}\circ x_0^{-1})=\\ = a \circ x_0 \circ (x_0^{-1} \circ a^{-1}\circ x_0^{-1})=\\ x_0^{-1}=(x_0^{-1} \circ a^{-1}\circ x_0^{-1}) \ast a. $$

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  • $\begingroup$ I was doing the calculations myself when you posted - got the exact same answer. Obrigado! $\endgroup$ – AJ44 Jan 21 '16 at 17:07
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I don't think you have anything yet.

1) You say G is non-empty because $x_0 \in G$. Well yes, but you know that as a set $<G, \circ> = <G, *>$ so $<G, *>$ is non-empty.

What we have to prove instead is that ever $g \in G$ can be written as $h \circ x_0 \circ i$. Which it can as $g = g \circ x \circ c_0^{-1}$

2) Before you could make the claim you had to show all $a = a' \circ x_0 \circ a~$, which we did above. So as $a,b$ in $<G, \circ$> then $a,b$ in $<G, *>$ so $a \circ x_0 \circ b \in <G, \circ>$ which as a set is $<G, *>$. But we did have to show that all a $in G$ were expressible as $a' \circ x_0 \circ a~$ first

3) $a * I_G = a \circ x_0 \circ I_G = a \circ x_0 \ne a$

So that won't do it! we need to find $e$ such that $c \circ x_0 e = c = e \circ x_0$ for all $c \in G$.

If $c \circ x_0 e = c$ then $e = x_0^{-1} \circ c^{-1} \circ c = x_0^{-1}$.

Meanwhile $c*x^{-1} = c \circ x_0 \circ x_0^{-1} = c = x_0^{-1} \circ x_0 \circ c $.

So $x^{-1}$ is the indentity element of $<G , *>$

3) Finally inverse: For $a \in <G, *>$ what $a \circ x_0 \circ b = b \circ x_0 \circ a = x^{-1}$?

That would be $x_0^{-1} \circ a^{-1} \circ x_0^{-1}$. Try it.

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