Group Theory Question on the set of real numbers

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All groups are defined by an operator. A common example used would be the set of real numbers under addition, in which case $\Bbb R$ is defined as a group under the operator $+$. And obviously the group axioms hold for $(\Bbb R, +): a+b \in \Bbb R, \forall a,b \in \Bbb R; (a+b)+c = a+(b+c), \forall a,b,c \in \Bbb R.$ The identity element is $0$ under addition and the additive inverse of $a$ is $-a$.

What the question wants is for you to use the * that they defined instead of using +.

So first you would show that G is closed under $*$. ie. $a*b \in G, \forall a, b \in G$. More specifically, for any real numbers $a \neq -1, b \neq -1$ (in other words $a,b \in G$) you would show that $a + b + ab \neq -1$ (you could do this with proof by contradiction: Let $a+b+ab = -1$ and show that this contradicts the fact that $a \neq -1, b \neq -1$). Then that would mean that G is closed under *.

Next you would go through the group axioms using the operator *. So, for instance, a group requires associativity: $$(a*b)*c = a*(b*c)$$

To show this you would just do the calculations using $*$ and any $a,b,c \in G$. For the LHS: $$(a*b) * c = (a + b + ab) * c = (a+b+ab) + c + (a+b+ab)c$$ Then show that this is equal to the RHS.

I'll show you how to find the identity element. Then it should be clear to you how to find the inverse element since you would follow the same procedure.

An identity element, $e$, requires that $a*e = e*a = a, \forall a \in G$. So first do the calculations using *: $$a*e = a + e + ae$$ $$e*a = e + a + ea$$

So what element in G should $e$ be equal to? Well, clearly $0$ is in G. And if $e = 0$ then we get: $$a*0 = a + 0 + 0 = a$$ $$0*a = 0 + a + 0 = a$$ Which means $0$ is the identity element in G.

The rest should be easy enough for you to figure out.

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Kara
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Kara

Senior at WSU. Math major, junior.

Updated on July 21, 2022

Comments

  • Kara
    Kara less than a minute

    Let G = {x $\in$ $\Bbb R$ | x $\neq$ -1} be the set of real numbers other than -1. For a, b $\in$ G, define
    a * b = a + b + ab.
    Prove that G is a group under the operation *.

    I know how to prove that something is a group, I'm just not sure what to do with the * operation. How do I prove things with that?

    • J126
      J126 over 8 years
      Do you know what it is that you must show and just can't do it? Or, do you not know what to show? What are the group axioms?
    • anon
      anon over 8 years
      Hint: if $a*b+1=(a+1)\cdot(b+1)$ then compare $(\Bbb R,*)$ with $(\Bbb R,+)$ in light of $x\leftrightarrow x+1$.
    • Hagen von Eitzen
      Hagen von Eitzen over 8 years
      @anon You meant: with $(\mathbb R^\times, \cdot)$.
    • anon
      anon over 8 years
      Oops yes. ${}{}$
    • Arthur
      Arthur over 8 years
      Kara, please do not edit your questions in an attempt to remove them.
  • andraiamatrix
    andraiamatrix over 8 years
    I was simply calculating (ab)*c. If you want, you can substitute $d = a*b$ so then you get $(a*b)*c = d*c$. This gives you $d*c = d + c + dc$ but we have from the definition of $*$ that $d = a*b = (a + b + ab)$ so substitute that wherever you had d before: $d*c = (a+b+ab) + c + (a+b+ab)c$. In order to show the LHS=RHS you would have to then multiply the c through everything in the final term and show that you get the same results with a*(bc). Just remember to do the calculations in the parentheses first.
  • andraiamatrix
    andraiamatrix over 8 years
    Recall in the definition of the inverse: $a*a^{-1} = 0$ since $0$ is your identity. So then you get $a + a^{-1} + aa^{-1} = 0$. This is an equation that you need to solve for $a^{-1}$. Just use high school algebra to isolate $a^{-1}$.
  • andraiamatrix
    andraiamatrix over 8 years
    You should not have got a number. Try solving again. I'll start you off: $a + a^{-1}(1+a) = 0$. Does that help?
  • andraiamatrix
    andraiamatrix over 8 years
    I'm not sure if you need to prove non-empty but if you want to you would have to look back at the definition of the group to check. Saying $a,b \in G$ is not sufficient. In this case, since the group is defined as real numbers $\neq -1$ then you can see its nonempty because $\Bbb R - \{-1\}$ is nonempty.