The question I'm having problems with involves proving the above groups are isometric. Therefore, I have to prove they are bijective (1-1, and onto) and homomorphic. I have done the group operations table for each and come up with:

Set $A = (\mathbb{Z}_4,+) = {0,1,2,3}$

Set $B = (\Bbb{Z}_5^*, \times) = {1,2,3,4}$

So $B$, can be mapped from $A$ with the function: $\alpha(x)=x+1$

We can see no element of $B$ is the image of more than one element in $A$, therefore we have proven a 1-1 correspondence, and onto.

Now I have to prove they are isomorphic by exhibiting a 1-1 corresponds α between their elements such that:

$a+b \equiv c\ (\text{mod}\ 4)$ if and only if $\alpha(a) \cdot \alpha(b)\equiv \alpha(c)(\text{mod}\ 5)$

I'm stuck here... do I just plug in all the possible values of a, b, and c? I suppose there should be 4 ways to do this...

As an example:

$a = 1, b = 2, c = 3$

$1 + 2 = 3(\text{mod}\ 4)$

$3\equiv 3(\text{mod}\ 4)$


$\alpha(1) \cdot \alpha(2) \equiv \alpha(3)(\text{mod}\ 5)$

$2\cdot 3 \equiv 4(\text{mod}\ 5)$

$6\equiv 4(\text{mod}\ 5)??????$

I feel like I'm missing a key concept here. Been watching a ton of videos, but I'm missing something!

  • 1
    $\begingroup$ The additive group $\Bbb{Z}_4$ is cyclic (generated by 1). So try to find a generator of $\Bbb{Z}_5^*$ (this is what will map to $1 \in \Bbb{Z}_4$). For example, $2$ generates $\Bbb{Z}_5^*$, since $2^0 = 1$, $2^1 = 2$, $2^2 = 4$, and $2^3 = 3$. You should be able to define $\alpha(0) = 1$, $\alpha(1) = 2$, $\alpha(2) = 4$, $\alpha(3) = 3$. $\endgroup$ – Nick Oct 6 at 3:02
  • $\begingroup$ are you sure about your question? cause in $\Bbb Z_5$ there is also element of 0... $\endgroup$ – friedvir Oct 6 at 3:08
  • 1
    $\begingroup$ @friedvir: multiplicative group $\Bbb Z_5^*$ doesn't contain $0$ $\endgroup$ – J. W. Tanner Oct 6 at 12:25
  • $\begingroup$ Cf. this question $\endgroup$ – J. W. Tanner Oct 6 at 13:21

Your error is right at the start, with the map $x \to x+1$. That is a bijection, but not a group isomorphism since it does not respect the operations in the group.

In your map $0$ must go to $1$ (can you see why?). Then try to find a place for $1$ to go. Once you decide on that, the group operations will force the rest of the map. So experiment.


Define your function as $$0\to 1$$

$$1\to 2$$

$$2\to 4$$

$$3\to 3$$

and everything works fine.

Not every one-to -ne function is an isomorphism.



Any group of order $4$ is isomorphic to $(\Bbb Z_4,+)$ or to the Klein Vierergruppe. In the latter, the square of any element is the identity.

  • $\begingroup$ I think the OP is looking for an explicit isomorphism. $\endgroup$ – Ethan Bolker Oct 6 at 12:51
  • $\begingroup$ @EthanBolker: such was already given in another answer; my point was isomorphism can be established non-explicitly $\endgroup$ – J. W. Tanner Oct 6 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.