Is Ker(f)={e} Enough to Prove Injectivity for a Group Homomorphism?

[quasar987]

I suspect it is quite simple and works by contradiction, but I can't see why for a group homomorphism f, ker(f) = {e} ==> f is injective. Any help?

 

[matt grime]

That's the definition isn't it? Oh, you mean f(x)=f(y) implies x=y... well, if f(x)=f(y) then what is f(xy^{-1})?

 

[AKG]

Suppose f is not injective, i.e. for x \neq y,\ f(x) = f(y). Then:

f(x)(f(y))^{-1} = e'

where e' is the identity of the codomain of f (e is the identity of the domain of f).

f(xy^{-1}) = e'

xy^{-1} \in \mathop{\rm Ker}(f)

Suppose xy^{-1} = e, then x = y, contradicting the assumption that they are unequal. So some element Ker(f) contains some element different from e, so it's not {e}.

 

[mathwonk]

its called "subtraction".

 

[quasar987]

Nice AKG.

(What is called "substraction"?)

 

[matt grime]

another unnecessary proof by contradiction. a good exercise for you would be to make it a shorter non-contradiction proof, quasar, it's a one liner.

mathwonk's hint is that proving A=B is the same as proving A-B=0, ie take it all onto one side. you presumalby know that a linear map M is injective if the only solution to Mx=0 is x=0, well, this is exactly the same, only the operation isn't addition/subtraction but general group composition.

 

[quasar987]

Meanwhile, there's another one I don't get:

f: G-->G' is an homomorphism. If H' is a normal subgroup of G', show that f^{-1}(H') is a normal subgroup of G.

I've shown that it's a subgroup, but can't show it's normal. What I tried:

If H' is normal, then for all g' in G', g'H' = H'g'. In particular, for g' 's such that there exists a in G such that g' = f(a), we have f(a)H' = H'f(a). This means that for all x in f^{-1}(H&#039;) and a in G, f(a)f(x) = f(x)f(a) <==> f(ax)=f(xa). But this does not imply that ax=xa unless f is injective. Inversely supposing that f^{-1}(H&#039;) is not normal leads to no (obvious) contradiction because it would means that there exists a* in G and x* in f^{-1}(H&#039;) such that a*x* \neq x*a*, which does not imply that f(a*x*) \neq f(x*a*).

 

[matt grime]

why would you want ax=xa anyway? that is not the condition of normality of a subgroup. H is normal if for gx in gH there is an element y of H such that yg=gx, there is no requirement that x=y.

i think it's quite straight forward, if H is the inerse image of H', then gx in H implies f(gx) \in f(gH)=f(g)H&#039;=H&#039;f(g) which implies that there is a y in H such that f(yg)=f(xg) so that, pulling back by f gx is in Hg too.

did you redo the proof of the last problem to be a one line proof NOT by contradiction? it would be helpful to your understanding to do it.

 

[quasar987]

why would you want ax=xa anyway? that is not the condition of normality of a subgroup. H is normal if for gx in gH there is an element y of H such that yg=gx, there is no requirement that x=y.

I hadn't realized that.

did you redo the proof of the last problem to be a one line proof NOT by contradiction? it would be helpful to your understanding to do it.

May I have a hint?

 

[AKG]

quasar it's very simple. You want to start with the antecedent, and in one line find a bunch of implications, so you have a chain like A --> B --> C --> ... ---> Z, where A is the statement that Ker(f) = {e}, and Z is the statement "f is injective." Do you know how to express the antecedent symbolically? Do you know how to express the consequent symbollically? If so, then it's just some easy symbol manipulation.

(A <--> B) --> (A' <--> B')

(A <--> B) is the symbolic statement that Ker(f) is trivial, and (A' <--> B') is the statement that f is injective, and follows immediately if you know that f is a homomorphism, and you know how to do basic group multiplication (which results in something that looks like subtraction).

 

[matt grime]

As with a lot of proofs by contradiction the fist line "assume that something is not..." is not necessary.

f injective iff ker(f)={e} ( the => is easy since it is a special case of the the general property of being injective: f(e)=e so if anything else satisfies f(g)=e thetn g=e and ker(f)={e})

f(x)=f(y) implies f(xy^{-1}=e, which, by hypothesis means xy^{-1}=e as we required to show.

so we don't prove it by assuming the opposite and getting a contradiction since the proof is actually direct. the same can be done for cantor's diagonal argument.

 

[quasar987]

Ooh, I see .

(I haven't seen multiplication yet AKG)

 

[mathwonk]

my calling it "subtraction" might make it sound trivial, but it is a thoroughly proven technique in mathematics to change every problem into the problem of proving something is zero. it is usually done by subtraction.

the reason is that proving something is zero is usually easier.

e.g. in the mean value theorem to show there is a point c where f'(c) = [f(b)-f(a)]/(b-a) one subtracts the function f(a) + (x-a)[f(b)-f(a)]/(b-a)
and applies the Rolle theorem, i.e. the case where we want f'(c) = 0, which is easier to prove.

or for checking a function ahs a maximum, we first take a derivative and see if it is zero (same argument really as rolles).

or for showing some surface is flat we define a "curvature" \and show it is zero.

or for showing all non zero functions in a given region have a complex log, we define the fundamental group of the region and show it is zero.

the list is endless.

 

[AKG]

Ooh, I see .

(I haven't seen multiplication yet AKG)

Sure you have. The basic group operation is often called multiplication unless otherwise specified. But if the use of the word "multiplication" confuses you, ignore that word and replace that part of my post with:

"...and you know how to do (the) basic group operation..."

 

[quasar987]

k, I thought you meant the 'product of groups', which is the subject of the next section of my book.

 

[quasar987]

i think it's quite straight forward, if H is the inerse image of H', then gx in H...

You probably meant "gx in gH".

...implies f(gx) \in f(gH)=f(g)H&#039;=H&#039;f(g) which implies that there is a y in H such that f(yg)=f(xg)

You probably meant "such that f(yg) = f(gx),

...so that, pulling back by f gx is in Hg too.

I've thought a lot but can't figure it out. What do you mean by "pull back by f" ?

 

[AKG]

Since H' is normal, G'/H' forms a group. Define a function f' : G --> G'/H' by:

f'(g) = f(g)H'

It is easy to check that f' is a homomorphism. What my book calls the First Isomorphism Theorem will tell you that the kernel of a homomorphism is a normal subgroup of the domain, and in particular, Ker(f') is a normal subgroup of G.

Ker(f') = {g in G : f'(g) = H'} = {g in G : f(g)H' = H'} = {g in G : f(g) in H'} = f^-1(H')

 

[mathwonk]

did you appreciate post 13? it is a general principle vastly more useful than this one question, or perhaps equally useful, if you can see the ocean in a drop.

 

[quasar987]

Sweet proof AKG, but unfortunately I haven't seen that First Isomorphism Theorem that you use and would like a proof using only what I know. If matt could just clarify what he meant that'd be great .

mathwonk, no I didn't appreciated it very much, I'm sorry. I was aware of that principle/trick but do not see how it could apply here. :-/

 

[AKG]

Although I used part of what is called the First Isomorphism Theorem, I used a simple part of if that's very easy to prove. You can prove for yourself that Ker(f) is normal for any homomorphism f. It's extremely easy to prove, so I won't give any hints. The theorem says other things, however, such as the fact that if f : G --> G' is a homomorphism, and K is its kernel, then G/K is isomorphic to f(G), and that the mapping h : G/K --> f(G) defined by:

h(xK) = f(x)

defines an isomorphism between the domain and co-domain. So you can do this proof using my approach if you wanted without referring to the theorem because the part that you'd use is something you can easily prove yourself.

As far as what matt's doing, he's saying that we know that f(gx) = f(yg) for some y in H. You know that you can write gx in the form y'g for some y' in G (y' = gxg^-1). Suppose y' is not in H, then f(y') is not in H', so:

f(gx) = f(y'g) = f(y')f(g)

but we know f(gx) = f(yg) = f(y)f(g)

So f(y')f(g) = f(y)f(g), f(y') = f(y), but f(y') is not in H', and f(y) is, which is impossible, so f(y') is in H', so y' is in H, and so gx = y'g in Hg, so gH is a subset of Hg. Similarly, you can show that Hg is in gH. You will have that the left and right cosets of H coincide, and hence you will know that H is normal. I might used a few extra lines above, since I also wasn't sure what matt was doing at first. It seems to me that there was no need to introduce a y, so maybe had an idea for a more efficient proof using it. An even simpler, one line proof:

x in H. If H is normal, then gxg^-1 should be in H. Well:

f(gxg^-1) = f(g)f(x)f(g)^-1 in f(g)H'f(g)^-1 = H'f(g)f(g)^-1 = H', so gxg^-1 is in H, so H is normal.

 

[matt grime]

to be honest what i wrote was the first thing that popped into my head so it probably wo'nt be very accurate (the corrections wer correct in quasar's post) or elegant and may need tweaking to work properly. i can't even remember my thought processes from working it out.

 

[matt grime]

actually it is a one liner

if H is the preimage of H'

f(gHg^{-1])=f(g)f(H)f(g^{-1}) = f(H) thus gHg^{-1]=H

if you don't like doing it with groups then just do it with h in H instead of all of H at once.

edit, sorry, just reread AKG's post where he proves this first.

 

[mathwonk]

a subgroup is normal if it is the kernel of a homomorphism, i.e. the things that go to zero.

if H' is normal in G' then G'/H' is a group, so H = f^(-1) (H') is the kernel of the composition

G-->G'-->G'/H'. qed.

another application of my basic principle. i.e. change what you are looking for into zero (the identity).
this is exactly AKG's "sweet proof" in post 17.

 

[quasar987]

I'll take the one liner, thank you.

 

[quasar987]

Suppose K is a group. Then K is an infinite cyclic group iff K is isomorphic to the integers under addition.

I've shown everything except that if K is isomorphic to Z, then K is cyclic. Anyone got an idea?

 

[shmoe]

You're trying to prove that a group isomorphic to the integers is cyclic? Are the integers cyclic?

 

[quasar987]

Mh, yeah it's that trivial huh?

In the book I'm using, there is a theorem stated before the one I'm trying to prove that says that 'Isomorphisms preserve all algebraic properties'. In particular, if f:G-->G' is an isomorphism, then G' is cyclic iff G is cyclic.

But I couldn't prove that theorem, because what are algebraic properties exactly?

 

[Muzza]

If x is a generator of G, then f(x) is a generator of G'.

The converse follows from the fact that f has an inverse.

 

[quasar987]

Yeah, but is there a way to prove the general statement "Isomorphisms preserve all algebraic properties" ?

 

[matt grime]

yes, since the algebraic properties of groups are by definition those things preserved by isomorphisms. isomorphic groups are identical as groups in terms of their behaviour as groups.

given a class of objects sharing some common definition an isomorphism is something that preserves exactly those properties that can eb deduced from the definition. that's why we study isomorphisms.

So, an iso of groups is a bijection between G and H such that f(xy)=f(x)f(y) and f(x^-1}=f(x)^{-1}. G and H are idenntical oas groups as the identification x <--> f(x) shows so all group theoretic (ie the algebraic properties of groups) are preserved by definition.

 

[quasar987]

If an algebraic property is by definition a property that is preserved by isomorphisms, isn't proving that all algebraic properties are preserved by isomorphisms futile?

And in identifying the algebraic properties of groups, one would have to test them one by one: Is the order preserved? Is the property of being cyclic preserved. If H is a normal subgroup in G, is f(H) normal in G' too?, etc.

 

[matt grime]

An isomorphism is exactly something that preserves the properties of a group that are a direct consequence of it being a group, or drectly are defined by group theoretic terms alone, and you can prove all of these very easily from the definition of an iso which is why they are good questions to set beginning students.

Let f be an iso from G to H

Examples.

if x is conjugate to y f(x) is conjugate to f(y). Proof: there is an z such that yz=zz hence f(y)f(z)=f(z)f(x)

if x has order n so does f(x)
G is abelian if and only if H is abelian

K is normal in G if and only if f(K) is normal in H

G is cyclic iff and only if H is

and so on and so on.

The definition of iso is straight forward, it is up to you it preserves properties of groups. those properties it preserves are exactly those things that we think of as being innately algebraic FOR GROUPS, that is the intrinsic properties of groups

 

[quasar987]

Concerning another problem:

If a is an element of a group G, there is always a homomorphism from Z to G which sends 1 to a. When is there a homomorphism from \mathbb{Z}_n to G which sends [1] to a? What are the homomorphisms from \mathbb{Z}_2 to \mathbb{Z}_6?

I answered that there is always such an homomorphism in the person of f([m]) = a^m. The homomorphisms from Z_2 to Z_6 are f_i([m]) = i[m], i = 0,1,...,5

This looked good to me though I was not entirely sure there exists no other homomorphisms from Z_2 to Z_6. But then the next question is

Suppose G is a group and g is an element of G, g \neq e. Under what conditions on g is there a homomorphism f : Z_7 --> G with
f([1]) = g ?

I would answer 'always', but it wouldn't make sense to ask that question if the answer to the general question 'When is there a homomorphism from \mathbb{Z}_n to G which sends [1] to a?' is 'always'.

What's wrong here?

 

[matt grime]

f is a homo. if x has order n the e=f(x^n)=f(x)^n so what is the order of f(x)? when you've figure that out redo that last post

 

[quasar987]

You've said it yourself in post #32: n. But how is this related to this problem ?!

 

[matt grime]

no, post 32 was about isomorphisms. think about it. if g is an element of any group and g^n=e what does that tell you about the order of g?

from what you;ve said you appera to think that there is a non-zero homomorphism from Z_7 to Z, and there isn't.

 

[quasar987]

The order of g divides n.

How is f([m]) = a^m not an homomorphism from Z_7 to Z?

 

[matt grime]

What is a? why does the order m, which divides 7, divide the order of a^m? how can it? there are no elements of finite order in Z apart from 0 are there? so the only homo from Z_7 to Z is the trivial one.

 

[matt grime]

one thing that would really help you is to learn the isomorphism theorems (and don'r sey they are in another chapter. one can prove all these things without tem but to understand why they are true it is ehlpful to know them).

if f is a homo from G to H it means that there is a quotient group of G isomorphically embedded in or has a copy inside or is a subgroup of H. Z_7 has no non-trivial quotients os it's clear that theer can be no isomorphic embedding of Z_7 in Z, or less fancilly, Z_7 is not a subgroup of Z, though you have stated it is.

 

[quasar987]

a is some element in Z.

Defined in this way, f([m]+[l]) = f([m]) f([l]), so f is an homomorphism. That's how I see it. How do you?

 

[Muzza]

But is it even a function if you don't place more restrictions on a? [8] = [1] (in Z_7), so f([8]) = f([1]), i.e a^8 = a (if f were a function)...

 

[matt grime]

oh come on! I've already explained it isn't a homomorphism unless a=0 at which point it is trivial. in the sepciifc example you gave you were asked to send 1 in Z_7 to g in G where g=/=e so that rules out the trivial case.


Let is use addition since Z is an additive group, as is Z_7

you are sending [m] in Z_7 to am in Z

this isn't a homomorphism, and you should prove it. HINT 3+4=0 in Z_7 and doesn't in Z

 

[quasar987]

matt, let's start with the first problem, which is

If a is an element of a group G, there is always a homomorphism from Z to G which sends 1 to a. When is there a homomorphism from Z_n to G which sends [1] to a? What are the homomorphisms from Z_2 to Z_6?

But is it even a function if you don't place more restrictions on a? [8] = [1] (in Z_7), so f([8]) = f([1]), i.e a^8 = a (if f were a function)...

Oh, so the problem is that f is not even a function! Interesting... Well it would be if a is e!

Ok, so now to the first question, I can answer that f, defined as

f([m])=a^m

(supposing G is multiplicative) is an homomorphism that sends [1] to a always if a=e and under the condition that n divides o(G) if a \neq e. That's that for this particular function. Now, are there other possible functions that are homomorphisms and that send [1] to a? I don't know.

To the other, I would say that the homomorphisms from Z_2 to Z_6 are

\{f_i([m]) = 3i[m]\}_{i \in \mathbb{Z}}

 

[matt grime]

you appear to have an infinitely family of homomorphisms, are you claiming that they are different?

Suppose the map from Z_2 to Z is an isomorphism, then it must send 0 to 0 and 1 to an element of order 2, and there is only one of those, 3. If it isn't an isomorphism then it must send 0 and 1 to 0 so there are two homos.

try it if we think of maps Z_3 to Z_6

 

[quasar987]

No, I don't claim that; the f_i's for odd i's all do the same thing, namely send [m] to [3] and the f_i's for even i's all do the same thing, namely send [m] to [0].

Let me now ponder on your isomorphism argument there.

 

[quasar987]

I can't crack the code

Suppose the map from Z_2 to Z is an isomorphism

The map? What map? Also, is it not impossible to have an isomorphism btw a group of order 2 and one of infinite order?

, then it must send 0 to 0 and 1 to an element of order 2, and there is only one of those, 3.

Please use the [] notation for elements of Z_n in order to avoid confusion, thx.

If it isn't an isomorphism then it must send 0 and 1 to 0 so there are two homos.

Two homos! Could you extrapolate the logical steps leading to that conclusion?

 

[matt grime]

apologies, i meant to say "suppose f is an injective homomorphism, ie its kernel is {e}, from Z_2 to Z_6" so there were at least 2 errors there.

no, i won;t use brackets for Z_n since it is unnecessary.

 

[quasar987]

Ok, so now to the first question, I can answer that f, defined as

f([m])=a^m

(supposing G is multiplicative) is an homomorphism that sends [1] to a always if a=e and under the condition that n divides o(G) if a \neq e. That's that for this particular function. Now, are there other possible functions that are homomorphisms and that send [1] to a? I don't know.

How do we show that all homomorphisms that send [1] to a are of the form above?

edit: Or rather, how do we show that f above is the only homo that sends [1] to a.

 

[AKG]

\mathbb{Z}_n is cyclic, so to find what a homomorphism does to any element m in Z_n, you only need to know what it does to 1, since 1 generates the group:

f(m) = f(1^m)
= (f(1))^m since f is a homomorphism
= a^m since we stipulated that f sends 1 to a

Note that normally, a^b means that we multiply a, b times. However, when dealing with groups, it is common to refer to the group operation as multiplication (regardless of what it is - addition, addition mod n, multiplication, composition, etc.) and it is common to use "multiplicative notation," regardless of what the operation is. So 1^m means that we do the group operation to 1, m times. The group operation happens to be addition modulo n, so we might prefer to write m(1) instead of 1^m, but as long as you keep in mind that there is only one operation that concerns us, then the notation 1^m meaning to add 1 to itself m times shouldn't confuse you.

Anyways, now you should be able to see that f must send m to a^m. There is only really one restriction that needs to be placed in order for f to be a function. Once this restriction is in place, not only will f be a function, but by the way we defined it, it will automatically be a homomorphism. Can you figure out this restriction? n need not divide the order of G. In fact, G could be an infinite group, in which case it may not even make sense to say whether n divides |G| or not.

 

[quasar987]

It must be that the order of a divides n: o(a)|n