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I am doing a tutorial question where it ask whether or not the groups of order 4 $(\mathbb{Z}_4,+\mod 4)$ and $(U(5),\times _5)$ are isomorphic to each other.

I drew out the Cayley tables and got

$$ \begin{array}{c|cccc} &0&1&2&3\\ \hline 0&0&1&2&3\\ 1&1&2&3&0\\ 2&2&3&0&1\\ 3&3&0&1&2 \end{array} \qquad \qquad \qquad \begin{array}{c|cccc} &1&2&3&4\\ \hline 1&1&2&3&4\\ 2&2&4&1&3\\ 3&3&1&4&2\\ 4&4&3&2&1 \end{array} $$

The answers said that they are isomorphic because they are both cyclic.

But looking at the Cayley tables I don't think you can define a bijection between elements of the groups (the diagonals do not match).

Can I get clarification on which interpretation is correct?

raynerk
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  • The function $\phi: \mathbb Z_4 \to U(5)$ which takes $0 \to 1, 1 \to 2, 2 \to 4, 3 \to 3$ seems like a decent candidate. – Gregory Sep 09 '23 at 01:32
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    They are isomorphic. Remember that on a bijection you have to rearrange the rows and columns. Try putting the $U(5)$ cayley table in the order $1 \ 2 \ 4 \ 3$ instead of $1 \ 2 \ 3 \ 4$. Do the tables match now? How might I have known that this order would work? Is there a way you can figure out the correct order yourself in future problems? What does this have to do with the concept of an isomorphism? – HallaSurvivor Sep 09 '23 at 01:32

4 Answers4

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See that $U(n)$ is the group of units of $\mathbb{Z}_n$ under multiplication modulo $n,$ i.e., $U(n)=(\mathbb{Z}_n^*,\times_n).$

We have a result regarding the cyclicity of $U(n),$ in the following article by David R. Guichard, “When Is $U(n)$ Cyclic? An Algebraic Approach,” Mathematics Magazine 72 (1999): 139–142.

$U(n)$ is cyclic if and only if $n$ is $1,$ $2,$ $4,$ $p^k,$ or $2p^k,$ where $p$ is an odd prime.

The answer to your question is a direct consequence of this result and the fact that there is only one cyclic group, upto isomorphism, of a given order.

MathRookie2204
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But looking at the Cayley tables I don't think you can define a bijection between elements of the groups (the diagonals do not match).

Just interchange the second table elements 3 and 4 [The row order in Cayley table is immaterial but you should keep same order in the column] $$. \begin{array}{c|cccc} &\color{red}0&\color{blue}1&\color{orange}2&3\\ \hline \color{red}0&\color{red}0&\color{blue}1&\color{orange}2&3\\ \color{blue}1&\color{blue}1&\color{orange}2&3&\color{red}0\\ \color{orange}2&\color{orange}2&3&\color{red}0&\color{blue}1\\ 3&3&\color{red}0&\color{blue}1&\color{orange}2 \end{array} \qquad \qquad \qquad \begin{array}{c|cccc} &\color{red}1&\color{blue}2&\color{orange}4&3\\ \hline \color{red}1&\color{red}1&\color{blue}2&\color{orange}4&3\\ \color{blue}2&\color{blue}2&\color{orange}4&3&\color{red}1\\ \color{orange}4&\color{orange}4&3&\color{red}1&\color{blue}2\\ 3&3&\color{red}1&\color{blue}2&\color{orange}4 \end{array} $$

Now look at the table and finish what you want..!

Chinnapparaj R
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You wanted all of them to map in an order... which is not necessary...

Look at the map $f:\mathbb{Z_4}\to U_5$, such that $f(0)=1, f(1)=2, f(2)=4, f(3)=3$ it is a bijection, and by watching the Cayley table itself you can conclude that this is an Isomorphism

Praveen Kumaran P
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Use the first Cayley table to "iteratively" compute: $$1, \space1+1, \space(1+1)+1, \space[(1+1)+1]+1$$ Same with the second to compute: $$2,\space 2.2,\space (2.2).2, \space [(2.2).2].2$$ You'll find out that no entries in the two sequences are groups' identities but the last ones. What does it mean?

citadel
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