# Proving isomorphism on additive group $(\Bbb{Z}_4,+)$ and multiplicative group $(\Bbb{Z}_5^*, \times)$

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:

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

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

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

and

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

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

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

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

• 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 $\alpha(0) = 1$, $\alpha(1) = 2$, $\alpha(2) = 4$, $\alpha(3) = 3$. – Nick Oct 6 at 3:02
• are you sure about your question? cause in $\Bbb Z_5$ there is also element of 0... – friedvir Oct 6 at 3:08
• @friedvir: multiplicative group $\Bbb Z_5^*$ doesn't contain $0$ – J. W. Tanner Oct 6 at 12:25
• – 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.

Hint:

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.

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