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?

  • $\begingroup$ The two groups don't have the same cardinality, therefore they are not isomorphic. $\endgroup$ – Arnaud Mortier Feb 26 '18 at 15:19
  • $\begingroup$ $\mathbb{Z}_4$ have 4 elements viz. 0,1, 2, 3 and $U_5$ also have 4 elements viz. 1, 2, 3, 4. $\endgroup$ – tatha Feb 26 '18 at 15:21
  • $\begingroup$ $U(5)=(\mathbb{Z}/5)^*$ has $\phi(5)=4$ elements. $\endgroup$ – Dietrich Burde Feb 26 '18 at 15:21
  • $\begingroup$ Sorry for me $U_5$ denotes the group of complex fifth roots of unity. You should define it if it is otherwise. $\endgroup$ – Arnaud Mortier Feb 26 '18 at 15:24

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.

  • $\begingroup$ 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? $\endgroup$ – tatha Feb 26 '18 at 15:24
  • $\begingroup$ What if we are not able to see that both are cyclic, then how? $\endgroup$ – tatha Feb 26 '18 at 15:27
  • $\begingroup$ 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 $2$. $\endgroup$ – Dietrich Burde Feb 26 '18 at 15:28
  • $\begingroup$ 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? $\endgroup$ – tatha Feb 26 '18 at 15:35
  • 1
    $\begingroup$ 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? $\endgroup$ – Dietrich Burde Feb 26 '18 at 16:10

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.

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