Question

show that any two groups of order 2 are isomorphic

Answer 1

Any groups of order two have two elements: the identity, and another element of order two that is not the identity. You can draw a multiplication table if you'd like, or however else you would like to show this.

Answer 2

I guess I was thinking about how I can write a proof, but it just says "show".

Answer 3

Well, most of the time when it says show they do want a proof. I'm not sure what the best way to go about proving this is. I guess construct two groups of order two and then construct the isomorphism between them? I dunno.

Answer 4

You can rely on the previously-proved theorem of the uniqueness of the identity, and then show that if the other element was of order greater than two the group would necessarily be of order greater than two as well... and obviously any two groups with the identity and an element of order two are isomorphic

Answer 5

yeh ty

Answer 6

you can exhibit the isomorphism directly
the identity of G goes to the identity of G' etc

Answer 7

of course more is true since any group of prime order is cyclic and therefore isomorphic to \(\mathbb{Z_p}\)

Answer 8

Do either of you have a good place to read about this topic? I am a little confused because I thought we needed a fuction to map one group to the other for it to be an isomorphism. So how can we say that two groups are isomorphic if we dont know the fuction?

Answer 9

you do know the isomorphism
put \(e, a\in G\) and \(e',a'\in G'\) and the map is \(\phi:G\to G'\) given by \(\phi(e)=e'\) and \(\phi(a)=a'\) is an isomorphism

Answer 10

then clearly \(\phi\) is both one to one and onto, and you can check that this is an isomorphism by showing that \(\phi(ea)=\phi(e)\phi(a)\) and that \(\phi(aa)=\phi(a)\phi(a)\) as required

Answer 11

if it is not clear let me know

Answer 12

or if you prefer you can show that \(G:=\{e,a\}\cong \mathbb{Z_2}:=\{0,1\}\)

Answer 13

So isomoprohic says they can be maped one to one and onto.

Answer 14

for a function from G to G' to be an isomorphism it must be one to one, onto, and a homomorphism

Answer 15

the function i defined above is clearly one to one, and onto right?

Answer 16

yes

Answer 17

so all that is left to show it that it is a homomorphism

Answer 18

but I was asked if two groups are isomorphic, I was told nothing about a function that maps them

Answer 19

true, so you make up the map

Answer 20

ok great I understand. I think... I might be back:)

Answer 21

what you have to do is to find the function that works
i think the problem is that in this case it is so easy it escapes the possible difficulties

Answer 22

I cant find a good lecture video on this anywhere

Answer 23

ahh i c

Answer 24

it might be better to start with a somewhat harder example, like showing any group of order 3 is isomorphic to \(\mathbb{Z_3}\)

Answer 25

you have essentially no choices for a group of order 2
and homo must send the identity to the identity

Answer 26

*any

Answer 27

So if two groups are isomorphic then they are the same thing? in a nut shell

Answer 28

well yes

Answer 29

you cannot say they are "equal" but they are the same thing, they just look different

Answer 30

right, they act the same

Answer 31

for example, you write \(G=\{e,a\}\) and i write \(\mathbb{Z_2}\) but there is not difference

Answer 32

and this is because they are defined as groups, not just sets?

Answer 33

or you write \(G=\{e, a, a^2\}\) and i write \(\{0,1,2\}\) but the actions are the same

Answer 34

yes the elements of the group are just some symbols, it is the action that is important

Answer 35

for a better exercise, see that the klien 4 group \(V:=\{e, a, b, ab\}\) is NOT isomorphic to \(\mathbb{Z_4}\)

Answer 36

klein 4 group \(a^2=b^2=e\) and \(ab=ba\)

Answer 37

I dont know what klien is, this is an intro to upper dev class.

Answer 38

o

Answer 39

right i just proved that if a^2 = e for all a then the group is abelian

Answer 40

there are two groups of order 4
\(\mathbb{Z_4} \) and klein 4

Answer 41

yes

Answer 42

one group of order 1, one of order 2, one of order 3, two of order 4, one of order 5 and only when you get to order 6 do you find one that is not abelian
but notice when i say "two of order 4" i must mean "up to isomorphism" because of course you can relabel the elements if you like

Answer 43

tyvm

Answer 44

A good way to think about isomorphic groups is to think about the structure. Basically, groups that are isomorphic have the same group structure. The difference between them is simply a relabeling of the elements. This will make even more sense once you look at automorphisms, which are isomorphisms from a group to itself. In automorphisms, the set stays the same, and the relationships between the elements stay the same, it is just basically a permutation of the elements. It's like a symmetry of the group. With isomorphisms in general, the difference is that the underlying set of the group doesn't necessarily stay the same, but you can show that each element basically plays the same role in the algebraic structure.

If you've had linear algebra, this is comparable to change of basis of a vector space. When you do a change of basis of a vector space, the actual vector space stays the same, you're just using a different basis. So, you end up with linear transformations that we call similar, because they are the same transformation with a different basis. Those similar linear transformations have some interesting properties. For example, they have the same determinants and eigenvalues. The relationship between similar linear transformations is very much like the relationship between isomorphic groups.

Hopefully some of that made sense :)

Answer 45

yes lots, I did much work with change of basis ty