5
$\begingroup$

Can anyone explain to me the subgroups of the Klein-4 group? I'm trying to view it this way:

I want some groups that are not empty and $ab^{-1} \in H$, where $H$ denotes the subgroups I am looking for.

So the Klein-4 group is as follows: $V_4 = \langle a,b : a^2 = b^2 = (ab)^2 = 1\rangle $. I'm a bit confused by the $\langle $ and $\rangle $ to denote a cyclic group but I sort of follow what's going on.

I guess the group itself and the trivial subgroup are there...But can you explain to me how to find the rest?

Possible answer:

Trivial subgoups $ \{\ e \}\ $ has order one. Order 2: $\{\ e,a \}\ , \{\ e,b \}\ , \{\ e,ab \}\ $ Order 4: The group itself. As the order of the subgroups must divide the order of the group, these must be all the subgroups.

$\endgroup$
  • $\begingroup$ $<$ and $>$ denotes the presentation of the group, in this case. $\endgroup$ – John Smith Mar 3 '14 at 15:33
3
$\begingroup$

Hints:

Lagrange's theorem says the only possible sizes of subgroups and orders of elements are $1,2,4$.

The identity element is one of the elements in each of the subgroups, and each element of order $2$ generates a subgroup of order $2$. Are there any elements of order $4$? By thinking about these ideas, you should see how to come up with your list.

The notation $\langle a,b,c,\ldots\mid\ldots \rangle$ is a special one for describing a presentation of a group. It is related to set notation but is not exactly the same.

The letters in the left half denote the supply of symbols that generate the group. Multiplying these symbols just involves using exponents and writing symbols next to each other. For example, $a\cdot b=ab$, $ab\cdot b=ab^2$ and so on. So don't get me wrong: the left half isn't a complete list of what's in the group, it's a list of symbols that generate the group.

The equations in the right half are rules that control the behavor of the multiplication. For example in the Klein $4$ group, one of the rules is that $b^2=1$, so in fact $ab\cdot b=ab^2=a1=a$. These are called relations for this group, since they relate products to each other and control the group operation.

$\endgroup$
  • $\begingroup$ Isn't Klein-4 order 2? $\endgroup$ – user3200098 Mar 3 '14 at 15:37
  • $\begingroup$ After some reading, I can see that it is order 4, but how come? I'm having problems seeing that it is order 4. $\endgroup$ – user3200098 Mar 3 '14 at 15:40
  • 1
    $\begingroup$ @user3200098: it has four elements, clearly. What's confusing about that? Are you confusing the order of the group with its exponent, by any chance? $\endgroup$ – tomasz Mar 3 '14 at 15:41
  • $\begingroup$ I realize that if $a$ and $b$ are strict numbers, then it has the elements: $a^2, b^2, (ab)^2$ and $1$. But my confusion stems from imagining $a$ and $b$ as any real numbers...And therefore the group could have endless elements...It works for $1$ and $-1$...I think I'm having a difficult time understand exactly what the klein-4 group is. $\endgroup$ – user3200098 Mar 3 '14 at 15:45
  • 1
    $\begingroup$ @user3200098 it's ok :) After a while these things become like breathing, and you see the details work out more or less automatically. I wish you luck learning about more groups. It's a worthwhile journey! $\endgroup$ – rschwieb Mar 4 '14 at 18:32
2
$\begingroup$

$G=\{e,a,b,ab\}$ the relation says that every nontrivial element has order $2$.

Thus,$H_1=\{e,a\}$,$H_2=\{e,b\}$,$H_3=\{e,ab\}$,$H_4=\{e\}$,$H_5=G$ are all subgroups.

The notation $<x,y,z..>$ means that group is generated by elements $x,y,z..$ if a group is generated by one elements then it is cylic.In that case, $G=<a,b>$ means every element of $G$ can be written in terms of $a$ and $b$ satisfying the given relation.

$\endgroup$

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.