A few examples of group homomorphism

[zcahana]

Homework Statement

The exercise is to find examples of various homomorphisms from/to various groups.

Those I'm having problems with are:

a. f: (Q,+) --> (Q^+,*) which is onto.
b. f: U20 --> Z64 which is 1-to-1.
c. f: Z30 --> S10 which is 1-to-1.

Homework Equations

The Attempt at a Solution

for a. , the only homomorphism i could think which send 0 to 1 is f(x) a^x,
apparently, f sends many elements of (Q,+) to R instead of (Q^+,*).

for b. , the only insight i have is that U20 has 8 elements, and Z64 has 8*8 elements ... not to deep, i know ... but maybe is?

for c. , no insights at all.



I would appreciate any clues given.
Thanks,
Zvi.

 

[Martin Rattigan]

To get b. and c. out of the way first (then we can come back to a. which is more interesting).

b. isn't possible if by U_{20} you mean the group of residues relatively prime to 20 under multiplication. This is because, with that meaning, U_{20}\cong Z_4\times Z_2 which is not cyclic (prove both, or at any rate the latter!), whereas any subgroup of a cyclic group (Z_{64} in this case) is cyclic (also prove!).

c. Does (12) commute with (345)? What then is the order of (12)(345). Then find an element of order 30 in S_{10}.

 

[Martin Rattigan]

a. is also not possible. Look at an element x\in (\mathbb{Q},+) such that f:x\mapsto 2\in(\mathbb{Q}^+,\times) (which must exist). What can you say about x/2\text{?}

 

[zcahana]

Ok,

For a. , I concluded that if f(x) = 2 then f(x/2) = sqrt(2) , with contradiction to that f(x/2) is (positive) rational.

For b. , proved.

For c. , I marked s = (1,2)(3,4,5)(6,7,8,9,10) , o(s) = 30
and for any x in Z30, f(x) = s^x such that IM(f) = <s> , and f(x) is well defined because it's not dependent of the number one chooses to represent the residue x.Thanks for your help, Martin ;)