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What I got is (please correct if anything is wrong)

Given 2 groups $\mathbb{Z}_4$ and $U_5$, proving isomorphism starts with proving there exist a bijection between 2 groups. Since there are same number of elements in them we can always construct a bijection.

But the mapping which preserves group structure should be bijective.

As both groups are cyclic, and $\mathbb{Z}_4$ is with addition operation and $U_5$ is with multiplication operation.

I get that if one generator is mapped to another which could be $1\rightarrow2$, mapping is complete.

Mapping function could be defined as $f(i) = a^i$

By this, required condition $f(a*b)=f(a)*f(b)$ is easily satisfied.

All this is understood intuitively. But how do I convey what generators are mapped and how do I prove there is bijection without multiplication or cayley table?

tatha
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2 Answers2

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Both groups have $4$ elements, and both groups are commutative. Furthermore all groups $U(p)$ are cyclic, see this duplicate. Hence $U(5)$ must be isomorphic to $\mathbb{Z}/4$. Mapping a generator to a generator yields an isomorphism.

Dietrich Burde
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  • If we were to prove from 2 conditions mentioned above, then how do I prove bijection. Also mapping is from addition operation to multiplication. Which can be summarized as $f(i)=a^i$. How to clearly formulate this expression and prove its bijection? – tatha Feb 26 '18 at 15:24
  • What if we are not able to see that both are cyclic, then how? – tatha Feb 26 '18 at 15:27
  • Then it is also easy. Since commutative groups of order $4$ are only $\mathbb{Z}/4$ and $\mathbb{Z}/2\times \mathbb{Z}/2$, we just need to exclude the second - which follows from the fact, that not all elements in $U(5)$ have order $1$ or $2$. – Dietrich Burde Feb 26 '18 at 15:28
  • Thing I can't understand is bijective mapping. Can you also show what mapping is between these groups. Or is product table only way to show mapping? – tatha Feb 26 '18 at 15:35
  • Indeed, check that $f(n)=a^n$ is a group isomorphism. Do we have $f(n+m)=f(n)f(m)$? – Dietrich Burde Feb 26 '18 at 15:59
  • yes that is satisfied but how is it one to one and onto? – tatha Feb 26 '18 at 16:02
  • also, it is not very clear which element is mapped to which using $f(n)=a^n$ – tatha Feb 26 '18 at 16:02
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    If $f(n)=1$, then $n=0$. Hence $f$ is injective. Every injective map between finite sets is also surjective. The map $f(n)$ clearly maps $0$ to $a^0=1$, $1$ to $a^1$, ..., until $p-1$ to $a^{p-1}$. So where is the difficulty? – Dietrich Burde Feb 26 '18 at 16:10
  • I'm new to this kind of maths and it appears that representation and language is half the work. I could see the mapping is bijective but writing that was not clear. Writing $\mathbb{Z}_4$ as set of powers of $a$ and then showing mapping between them to ${0, 1, 2...}$ is articulate enough. But one question in context of $\mathbb{Z}_4$ and $U_5$, what exactly is $a$? Is it 1? – tatha Feb 26 '18 at 16:25
  • The cyclic group $C_p={e,a,a^2,\ldots ,a^{p-1}}$ means that $a$ is a group element, not a number. Like for polynomials, $x^2+3x+2$ does not mean that $x=1$. – Dietrich Burde Feb 26 '18 at 16:32
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To explicitly give an isomorphism:

Let $\phi: \mathbb{Z}_4\rightarrow\mathbb{Z}_5^\times$ such that $\phi(n)=2^n$. Then $\phi(a+b)=2^{a+b}=2^a2^b=\phi(a)\phi(b)$.

  • Assume $\phi(n)=1$, then $2^n\equiv_51$ so $n\equiv_40$, since $2$ is a generator for $\mathbb{Z}_5^\times$. $\phi$ is injective.
  • If $a\in\mathbb{Z}_5^\times$ then $a=2^m$, so $\phi(m)=a$. $\phi$ is surjective.
cansomeonehelpmeout
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