Hi,

Could you help me with this one:

Show that the group of integers under addition is isomorphic to the group of 2by2 matrices (with elements 1 (1st row, 1st column), n, 0, 1) under matrix multiplication, where n is an integer)

Thanks

Kerwin

and define $\phi$ thus:

${\displaystyle \phi (n)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill n\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)}$

Now you just have to check that this is an isomorphism.

Thanks a lot.

I also came across inner auto morphism... I don't quite understand this! (unfortunately my school library is quite deficient in books on this topic!)

So, for instance,

how do i prove that f is an isomorphism on (G,*) to (G,*) if f(x) = g^-1 xg, for g fixed belonging to (G,*)

Thanks a lot

We can prove f(x)f(y)=f(xy):

f(x)f(y)=g^-1 xgg^-1 yg=g^-1 xyg=f(xy),

and, if f(x)=f(y), we have g^-1 xg=g^-1 yg. Hence x=gg^-1 xgg^-1 =gg^-1 ygg^-1 =y.

Thus it is a one-one mapping. Hence f is an automorphism of G.

Kerwin

Hi,

for the first question I asked,

how would you go about showing the function is a bijection (one-one)...this may be trivial but i'm not sure how to write this up properly ! Also for the second question, Kerwin, is that sufficient to show f is a bijection! I think you've only shown it's an injection!

Thanks

To also show surjectivity you want to show that any y arises as f(x) for some x.
Think about gyg^-1 ... why does this always exist?

There are (at least) two more general aspects that you might want to think about. The following applies to functions between sets generally, not just isomorphisms between groups.

1- For a function from a finite set to itself (or indeed from one set to another of the same finite size), either of injectivity or surjectivity alone is actually enough for the function to be one-to-one. Can you see this?
Does this remain true for infinite sets? What do you think?

2- You will have noticed that h(y)=gxg^-1 is the INVERSE of f, and an invertible function is always a bijection.
IMPORTANT: the "inverse property" needs to work either way, so if you apply f first and then f-inverse you must get back to where you started, AND if you apply the inverse first followed by f you must get back to where you started too!

Can you find two functions between two sets such that they invert each other only in one direction but not the other?

(Don't worry if you find this hard. It's the kind of thing you are asked to come up with in your first term at university.)

Thanks Barakat,

I'll start working on your problems... which seem very interesting! To be honest, I find groups quite intriguing... especially sporadic groups which i'm realising are analogous to prime numbers, in the sense that they are essentially building blocks!

Anyway, got another question, here:

Let a be in group G
If element a has order n and element a^-1 has order m, then prove m = n.

Now, this is what I did:

we know aⁿ = e thus a^-n = e, thus (a^-1 )ⁿ = e, thus n = m.

I don't think this is rigorous enough! could you help?

Thank you very much...

PS. I'm only 15, but decided to do maths at uni...so I'm trying to understand all of this!

It's not just the sporadic groups which are the 'building blocks' - its really all the simple groups, the sporadic groups being just special cases of these.

Your proof above is nearly correct. At the end you're assuming that because (a^-1 )ⁿ = e then a^-1 has order n. This isn't necessarily true - all you've actually shown is that the order of a^-1 divides n. Now you need to repeat the argument to show that n divides m, and you're done.

James.

That's fine, providing you are convinced that a^-n = (aⁿ )^-1

Alternatively consider (aⁿ ).(a^-1 )^m =e.e=e and then cancel out pairs a.a^-1 from the middle of the left hand side, either m=n or you are left with a^k =e or (a^-1 )^k =e for a positive k less than the order of that element.

By the way,

if you have applied f on G to G, f(x) = x² and suppose f is a homomorphism how does this imply that G is abelian?

thanks

Because for all x,y in G,
x² y² = f(x)f(y) = f(xy) = (xy)² = xyxy

Now apply x^-1 to the LHS and y^-1 to the RHS.

Richard

Here's an isomorphism question:

If you have applied f on G to G, f(x) = x^-1 and suppose f is an isomorphism how does this imply that G is abelian?

Thanks

Just plug the definition of f into f(xy)=f(x)f(y) and note that (xy)^-1 = y^-1 x^-1 to conclude that any two inverses commute. Can you see why this is enough for G to be abelian?
(Hint: [x^-1 ]^-1 )