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

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In summary: But then, why is it necessary to express the antecedent and the consequent symbolically?In summary, the proof that f is injective if Ker(f) = {e} can be done directly without assuming the opposite and getting a contradiction, as f is injective if and only if Ker(f) = {e}. This can be shown by expressing the antecedent (Ker(f) = {e}) and the consequent (f is injective) symbolically and using basic group multiplication. The use of subtraction in mathematics is a proven technique to simplify proving something is zero, making it a common method in various mathematical proofs.
  • #1
quasar987
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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?
 
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  • #2
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})?
 
  • #3
Suppose f is not injective, i.e. for [itex]x \neq y,\ f(x) = f(y)[/itex]. Then:

[tex]f(x)(f(y))^{-1} = e'[/tex]

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

[tex]f(xy^{-1}) = e'[/tex]

[tex]xy^{-1} \in \mathop{\rm Ker}(f)[/tex]

Suppose [itex]xy^{-1} = e[/itex], then [itex]x = y[/itex], contradicting the assumption that they are unequal. So some element Ker(f) contains some element different from e, so it's not {e}.
 
  • #4
its called "subtraction".
 
  • #5
Nice AKG. :smile:

(What is called "substraction"?)
 
  • #6
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.
 
  • #7
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 [itex]f^{-1}(H')[/itex] 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 [itex]f^{-1}(H')[/itex] 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 [itex]f^{-1}(H')[/itex] is not normal leads to no (obvious) contradiction because it would means that there exists a* in G and x* in [itex]f^{-1}(H')[/itex] such that a*x* [itex]\neq[/itex] x*a*, which does not imply that f(a*x*) [itex]\neq[/itex] f(x*a*).
 
  • #8
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 [tex]f(gx) \in f(gH)=f(g)H'=H'f(g)[/tex] 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.
 
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  • #9
matt grime said:
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.

matt grime said:
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?
 
  • #10
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).
 
  • #11
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.
 
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  • #12
Ooh, I see :smile:.

(I haven't seen multiplication yet AKG)
 
  • #13
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.
 
  • #14
quasar987 said:
Ooh, I see :smile:.

(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..."
 
  • #15
k, I thought you meant the 'product of groups', which is the subject of the next section of my book.
 
  • #16
matt grime said:
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".


matt grime said:
...implies [tex]f(gx) \in f(gH)=f(g)H'=H'f(g)[/tex] which implies that there is a y in H such that f(yg)=f(xg)

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

matt grime said:
...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" ?
 
  • #17
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')
 
  • #18
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.
 
  • #19
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 . :smile:

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. :-/
 
  • #20
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.
 
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  • #21
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.
 
  • #22
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.
 
  • #23
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.
 
  • #24
I'll take the one liner, thank you. :wink:
 
  • #25
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?
 
  • #26
You're trying to prove that a group isomorphic to the integers is cyclic? Are the integers cyclic?
 
  • #27
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?
 
  • #28
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.
 
  • #29
Yeah, but is there a way to prove the general statement "Isomorphisms preserve all algebraic properties" ?
 
  • #30
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.
 
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  • #31
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.
 
  • #32
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
 
  • #33
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 [itex]\mathbb{Z}_n[/itex] to G which sends [1] to a? What are the homomorphisms from [itex]\mathbb{Z}_2[/itex] to [itex]\mathbb{Z}_6[/itex]?

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 [itex]f_i([m]) = i[m][/itex], [itex]i = 0,1,...,5[/itex]

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 [itex]\neq[/itex] 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 [itex]\mathbb{Z}_n[/itex] to G which sends [1] to a?' is 'always'.

What's wrong here?
 
  • #34
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
 
  • #35
You've said it yourself in post #32: n. But how is this related to this problem ?! :confused:
 
<h2>1. Why is a group homomorphism important in mathematics?</h2><p>A group homomorphism is important in mathematics because it allows us to study the structure and relationships between different groups. By understanding how group homomorphisms work, we can better understand the properties and behaviors of groups, which are fundamental objects in abstract algebra and other areas of mathematics.</p><h2>2. What is the definition of a group homomorphism?</h2><p>A group homomorphism is a function between two groups that preserves the group structure. In other words, it maps elements from one group to elements in another group in a way that respects the group operations. This means that the result of combining two elements in the first group will be the same as combining their corresponding images in the second group.</p><h2>3. How is a group homomorphism different from an isomorphism?</h2><p>A group homomorphism is a function that preserves the group structure, while an isomorphism is a bijective homomorphism. This means that an isomorphism not only preserves the group operations, but also preserves the group's identity and inverses. In other words, an isomorphism is a one-to-one mapping between two groups that also preserves their algebraic properties.</p><h2>4. Can a group homomorphism be surjective but not injective?</h2><p>Yes, a group homomorphism can be surjective but not injective. This means that the function maps every element in the first group to an element in the second group, but there may be multiple elements in the first group that map to the same element in the second group. In other words, the function is "onto" but not "one-to-one".</p><h2>5. How is a group homomorphism related to other mathematical concepts?</h2><p>A group homomorphism is closely related to other mathematical concepts such as group actions, group representations, and ring homomorphisms. Group actions are similar to group homomorphisms in that they also preserve the group structure, but they act on a set rather than mapping between groups. Group representations are a way of studying groups by representing their elements as matrices or linear transformations, and group homomorphisms can be used to compare and relate different group representations. Ring homomorphisms are similar to group homomorphisms, but they preserve the structure of rings instead of groups.</p>

1. Why is a group homomorphism important in mathematics?

A group homomorphism is important in mathematics because it allows us to study the structure and relationships between different groups. By understanding how group homomorphisms work, we can better understand the properties and behaviors of groups, which are fundamental objects in abstract algebra and other areas of mathematics.

2. What is the definition of a group homomorphism?

A group homomorphism is a function between two groups that preserves the group structure. In other words, it maps elements from one group to elements in another group in a way that respects the group operations. This means that the result of combining two elements in the first group will be the same as combining their corresponding images in the second group.

3. How is a group homomorphism different from an isomorphism?

A group homomorphism is a function that preserves the group structure, while an isomorphism is a bijective homomorphism. This means that an isomorphism not only preserves the group operations, but also preserves the group's identity and inverses. In other words, an isomorphism is a one-to-one mapping between two groups that also preserves their algebraic properties.

4. Can a group homomorphism be surjective but not injective?

Yes, a group homomorphism can be surjective but not injective. This means that the function maps every element in the first group to an element in the second group, but there may be multiple elements in the first group that map to the same element in the second group. In other words, the function is "onto" but not "one-to-one".

5. How is a group homomorphism related to other mathematical concepts?

A group homomorphism is closely related to other mathematical concepts such as group actions, group representations, and ring homomorphisms. Group actions are similar to group homomorphisms in that they also preserve the group structure, but they act on a set rather than mapping between groups. Group representations are a way of studying groups by representing their elements as matrices or linear transformations, and group homomorphisms can be used to compare and relate different group representations. Ring homomorphisms are similar to group homomorphisms, but they preserve the structure of rings instead of groups.

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