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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!

Rick
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Drew Heasman
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    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 06 '19 at 03:02
  • are you sure about your question? cause in $\Bbb Z_5$ there is also element of 0... – friedvir Oct 06 '19 at 03:08
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    @friedvir: multiplicative group $\Bbb Z_5^*$ doesn't contain $0$ – J. W. Tanner Oct 06 '19 at 12:25
  • Cf. this question – J. W. Tanner Oct 06 '19 at 13:21

4 Answers4

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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.

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

$$1\to 2$$

$$2\to 4$$

$$3\to 3$$

and everything works fine.

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

J. W. Tanner
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Mohammad Riazi-Kermani
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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.

J. W. Tanner
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Let $\alpha(.)$ be the mapping from $(\mathbb{Z_4},+)$ to $(\mathbb{Z^*_5},\times)$

$\alpha(0+0) = \alpha(0) \times \alpha(0) \implies \alpha(0) = 1 $

Therefore $0$ maps to 1

Let $\alpha(1) = a, \alpha(2) = b \implies a^2 = b \implies a = 2 $ and $b = 4$

Therefore 1 maps to 2 and 2 maps to 4. Similar operations for 3 = 2 + 1 will show that 3 maps to 3.

Smit Sparsh
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