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Studying for group theory, and I ran in to two problems, stated as follows:

Question 1: To find the homomorphism of $\Phi:\mathbb{R^*} \rightarrow \mathbb{R}$ such that $\Phi(2)=3$, can we just consider $\Phi(a)=a$? So that when we have $\Phi(1*2)=(1+2)$ as given by the operation allowed on the real numbers, given by the mapping? ($\mathbb{R^*}$ multiplication on real numbers, $\mathbb{R}$ addition on real numbers)

Question 2: How can I show that when there are two finite subgroups $H_1$ and $H_2$ of a group $G$, then $H_1 \cap H_2$ have only one element, under the condition $|H_1|$ and $|H_2|$ are coprime?

I think I have to use Lagrange's theorem, and the obvious element seems to be the identity or I might be completely off. Can anyone help me show and understand how to approach question 2?

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  • $\begingroup$ If $\Phi(2)=3$, then I don't think we can "consider $\Phi(a)=a$". $\endgroup$ – Lord Shark the Unknown Nov 5 '17 at 7:29
  • $\begingroup$ TBH I dont know what kind of comment this is, its neither helpful or unhelpful. $\endgroup$ – Aurora Borealis Nov 5 '17 at 12:30
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Question 1: Since $\Phi$ is a homomorphism, $\Phi(a\cdot b)=\Phi(a)+\Phi(b)$ for all $a,b\in \mathbb{R}^{*}$. Homomorphisms send the identity to the identity, so $\Phi(1)=0$. These two properties remind us of $\log$. Consider $$\Phi(x)=\frac{3\log|x|}{\log 2}\,.$$ I leave it to you to verify that this a homomorphism satsifying the property $\Phi(2)=3$.

Question 2: Take any $x\in H_1\cap H_2$. Lagrange's Theorem tells that the order of $x$ divides $|H_1|$ and $|H_2|$. Since $|H_1|$ and $|H_2|$ are coprime, their only common factor is $1$. Thus, the order of $x$ is $1$, so $x=e$.

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  • $\begingroup$ Thank you very much, yes I will try to understand it. $\endgroup$ – Aurora Borealis Nov 5 '17 at 12:31

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