Let $G$ be the real numbers under addition and $G'$ the positive real numbers under multiplication. Prove that $G$ and $G'$ are isomorphic under the mapping $x\phi = 2^x$
Prove that $G$ and $G'$ are isomorphic
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abstract-algebra
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13Well you have the isomorphism, just prove that $\phi$ is group homomorphism and bijective... – 2011-11-09
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1The homomorphism property is just a law of exponents. You can show the function is bijective by demonstrating there is an inverse (which is $\mathrm{log}_2(y)$). – 2011-11-09
1 Answers
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These things might be of some use.
Show that $\phi$ satisfies: $\phi(x+y) = \phi(x) \cdot \phi(y)$
Show that if $\phi(x) = \phi(y)$, then $x =y$.
And for showing $\phi$ is onto note that for any $x > 0 \in \mathbb{R}^{\ast}$ $\exists$ $y =(\: \log_{2}(x)\: ) \in \mathbb{R}$ such that $\phi(y) = x$.
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5Argh; we're dealing with a group homomorphism. Don't check injectivity by verifying $\phi(x)=\phi(y)\Rightarrow x=y$; check injectivity by verifying $\phi(x)=1\Rightarrow x=0$. – 2011-11-09
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0@ArturoMagidin: Is there anything wrong in my argument. $2^{x}=2^{y} \Rightarrow 2^{x-y}= 1 \Rightarrow x=y$. I wanted to prove injectivity in this way. I don't really know why you pinpointed the above comment. – 2011-11-09
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0In *most* cases it is more advisable to check the kernel of a group homorphism than to do check that $\phi(x)=\phi(y) \Rightarrow x=y$. But in this case I guess it's all the same. – 2011-11-09
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1@Chandrasekhar: There is nothing inherently wrong; it's just I get frustrated with students who try to check injectivity the way you suggest, when the extra structure provides such a useful short-cut. Of *course* you can always prove injectivity by showing $\phi(x)=\phi(y)\Rightarrow \phi(x)\phi(y)^{-1}=1\Rightarrow \phi(xy^{-1})=1$ and then showing $xy^{-1}=1$, but then why bother proving the theorem characterising injectivity in terms of the kernel if the first place? – 2011-11-09