0

I can write $K_4=\{e,a,b,ab\}$ and $Z_2 . Z_2=\{(0,0),(0,1),(1,0),(1,1)\}$ How do I relate the two to show an ismorphic relation.
I also know that each element has order 2. How does that help ?

Diyanko Bhowmik
  • 13
  • Hint: It can be very helpful to identify some (minimal) generating sets. As any isomorphism must map generating sets to generating sets. – Justin Benfield Apr 30 '17 at 21:19

4 Answers4

2

Any two finite abelian groups having the same number of elements of each order are isomorphic.

Here,both $K_4$ and $Z_2 \times Z_2$ are abelian groups both having $3$ elements of order $2$ & an identity.

Styles
  • 3,609
  • If true, that seems highly non-obvious to me. – Chessanator Apr 30 '17 at 20:17
  • RossMillikan and Chessanator: Apology,I mistyped the statement. – Styles Apr 30 '17 at 20:19
  • @RossMillikan:What is the fallacy now? – Styles Apr 30 '17 at 20:27
  • 2
    Now I suspect it is true, but it is a non-trivial result that OP probably does not have available. The point of the question is to demonstrate an isomorphism. I believe there is a theorem that all abelian groups are direct products of cyclic groups and you can then match up the factors. – Ross Millikan Apr 30 '17 at 20:34
  • @RossMillikan:https://math.stackexchange.com/questions/1296833/if-i-know-the-order-of-every-element-in-a-group-do-i-know-the-group – Styles Apr 30 '17 at 20:39
1

Try constructing the bijection between them. Since the identity has to be preserved, $e$ must be sent to $(0,0)$. Send $a$ to any of the other elements. Send $b$ to another. Since we need a bijection, $ab$ must be sent to the final element.

All you have to do now is check that what you construct is a group homomorphism. (You only have finitely many things to check).

Chessanator
  • 5,673
  • Can you help in creating the homomorphism? I am slightly puzzled. – Diyanko Bhowmik May 01 '17 at 17:17
  • Pretty much any bijective function you construct by the above method is a homomorphism. You only need to check that the function meets the definition of a homomorphism (Do you know that definition?) – Chessanator May 01 '17 at 17:42
1

Write $K_4=\left\{e,a,b,c\right\}$ where $e$ is the neutral element, $a^2=b^2=e$ and $ab=c$. (This completely determines the group).

Define $f:K_4\rightarrow \mathbb{Z}_2\times \mathbb{Z}_2$ by $$f(a)=(1,0) \mbox{ and } f(b)=(0,1).$$

It's easy to check that $f$ extends uniquely to a well-defined group morphism. Obviously $f(e)=(0,0)$ and $f(c)=f(a)+f(b)=(1,1)$. Thus $f$ is a bijection.

Mathematician 42
  • 13,176
0

There are $24$ bijections between $K_4$ and $Z_2 \times Z_2$ as sets. If you use the extra information that $e$ must map to $(0,0)$ that leaves only $6$ maps. If I'm not mistaken, all $6$ happen to be isomorphisms in this case. Which means you just need to find some bijection sending $e$ to $(0,0)$ and show that it's a homomorphism.

Sera Gunn
  • 27,981