How Can Group Homomorphisms Prove Injectivity and Surjectivity?

[Math Amateur]

Can someone please help me with the following problem - it should be simple but I have so far made no significant progress:

Given the following:

(i) \(\displaystyle g \ : \ B \to C \) and \(\displaystyle h \ : \ C \to B \) are group homomorphisms

(ii) \(\displaystyle g \circ h = id_C \) where \(\displaystyle id_C \) is the identity mapping on C

Show the following:

(i) g is injective

(ii) h is surjective

I would appreciate some help.

Peter

 

[Nono713]

It might help to draw a diagram. Suppose $B$ has three elements, and $C$ has five. Draw some $g$ and $h$ to try and make $g \circ h$ the identity mapping on $C$, do you see the problem? What if $B$ had more elements than $C$, does it work then? What can you deduce?

Formal hint: If $g \circ h$ is the identity mapping then it is a bijection, what does that say about $g$ and $h$?

 

[caffeinemachine]

Can someone please help me with the following problem - it should be simple but I have so far made no significant progress:

Given the following:

(i) \(\displaystyle g \ : \ B \to C \) and \(\displaystyle h \ : \ C \to B \) are group homomorphisms

(ii) \(\displaystyle g \circ h = id_C \) where \(\displaystyle id_C \) is the identity mapping on C

Show the following:

(i) g is injective

(ii) h is surjective

I would appreciate some help.

Peter

I think I have a counterexample.

Let $B$ be a finite group with a subgroup $H$ such that $|G|=2$ and $|H|>1$. (Such groups exist).

Let $b\in B$ be an element of order $2$ in $B$. (Such a $b$ exists by Cauchy's theorem).

Define $C=\{1,-1\}$ under multiplication.

Define $g:B\to C$ as $g(h)=1$ for all $h\in H$ and $g( bh )=-1$ for all $h\in H$.

It can be seen that $g$ is a homomorphism.

Define $h:C\to B$ as $h(1)=e_B$ and $h(-1)=b$.

It's is easily seen that $h$ too is a homomorphism.

Note that $g(h(1))=g(e)=1$ and $g(h(-1))=g(b)=-1$.

Thus $g\circ h=id_C$. But neither $g$ is injective and nor $h$ is surjective.

 

[Math Amateur]

It might help to draw a diagram. Suppose $B$ has three elements, and $C$ has five. Draw some $g$ and $h$ to try and make $g \circ h$ the identity mapping on $C$, do you see the problem? What if $B$ had more elements than $C$, does it work then? What can you deduce?

Formal hint: If $g \circ h$ is the identity mapping then it is a bijection, what does that say about $g$ and $h$?

Thanks Bacterius.

I took your advice and learned something straight away ... but can someone check that my analysis is correct so far ...

I took as an example case:

\(\displaystyle B = \{ b_1, b_2, b_3 \} \)

and

\(\displaystyle c = \{ c_1, c_2, c_3, c_4, c_5 \} \)

I created an example mapping h as follows: (see attachment)

\(\displaystyle h(c_1) = b_2 \)
\(\displaystyle h(c_2) = b_1 \)
\(\displaystyle h(c_3) = b_3 \)
\(\displaystyle h(c_4) = b_1 \)
\(\displaystyle h(c_5) = b_3 \)

Now, of course, since we have more elements in C, some of the elements of C must map at least twice twice onto the same element of B, making it impossible to define a function \(\displaystyle g \ : \ B \to C \) as a function that results in \(\displaystyle g \circ h = id_C \) since this would require:

\(\displaystyle g(b_1) = g(h(c_2) = c_2 \)

and also

\(\displaystyle g(b_1) = g(h(c_4) = c_4 \)

which violates the definition of a function let alone not resulting in an identity mapping!

BUT ... then it seems that, for a start, anyway h has to be injective ?? which is not what I was seeking to show ...

It is starting to look as if both g and h have to be bijections and also that they would have to be inverses of each other ...

Can someone please help further ...

Peter

 

[Nono713]

There's something wrong with the problem. Are you sure it's not $h$ that's meant to be injective and $g$ that should be surjective? Otherwise I don't see how $g \circ h$ could be the identity mapping, as $h$ not being injective would imply there exist distinct $c_1$, $c_2$ such that $h(c_1) = h(c_2)$, i.e. $(g \circ h)(c_1) = (g \circ h)(c_2)$, and so $g \circ h$ cannot be the identity mapping.

Though admittedly I'm thinking that the fact that they are group homomorphisms should probably come into play at some point, because so far the reasoning applies to functions defined on sets in general and not specifically groups. But I'm not seeing it, because as far as I can tell your analysis is correct and for the problem's conditions to be met $h$ and $g$ must be bijections inverses of one another.

 

[Math Amateur]

There's something wrong with the problem. Are you sure it's not $h$ that's meant to be injective and $g$ that should be surjective? Otherwise I don't see how $g \circ h$ could be the identity mapping, as $h$ not being injective would imply there exist distinct $c_1$, $c_2$ such that $h(c_1) = h(c_2)$, i.e. $(g \circ h)(c_1) = (g \circ h)(c_2)$, and so $g \circ h$ cannot be the identity mapping.

Though admittedly I'm thinking that the fact that they are group homomorphisms should probably come into play at some point, because so far the reasoning applies to functions defined on sets in general and not specifically groups. But I'm not seeing it, because as far as I can tell your analysis is correct and for the problem's conditions to be met $h$ and $g$ must be bijections inverses of one another.

Thank you Bacterius.

You write:

"Are you sure it's not $h$ that's meant to be injective and $g$ that should be surjective?"

Yes, I was beginning to think that myself ... ...

The problem comes from a post Deveno sent me regarding a problem on split short exact sequences - see post at http://mathhelpboards.com/linear-abstract-algebra-14/exact-sequences-split-sequences-splitting-homomorphisms-9466.html

In that post, Deveno wrote:

" ... ... There is a "hidden" observation underlying this:

If $g \circ h$ is bijective, then $g$ is injective, and $h$ is surjective. Before you go any further, convince yourself this is true."

Given Deveno's knowledge of algebra I am inclined to take what he writes extremely seriously ... but perhaps this is a typo ...

So, yes ... I am beginning to think that he meant to say that $h$ is injective and $g$ is surjective.

Peter

Note:

(1) To clarify things a bit, in the exact sequences post \(\displaystyle g \circ h \) was given as the identity map on C and hence could be referred to as a bijection

 

[Math Amateur]

Thank you Bacterius.

You write:

"Are you sure it's not $h$ that's meant to be injective and $g$ that should be surjective?"

Yes, I was beginning to think that myself ... ...

The problem comes from a post Deveno sent me regarding a problem on split short exact sequences - see post at http://mathhelpboards.com/linear-abstract-algebra-14/exact-sequences-split-sequences-splitting-homomorphisms-9466.html

In that post, Deveno wrote:

" ... ... There is a "hidden" observation underlying this:

If $g \circ h$ is bijective, then $g$ is injective, and $h$ is surjective. Before you go any further, convince yourself this is true."

Given Deveno's knowledge of algebra I am inclined to take what he writes extremely seriously ... but perhaps this is a typo ...

So, yes ... I am beginning to think that he meant to say that $h$ is injective and $g$ is surjective.

Peter

Note:

(1) To clarify things a bit, in the exact sequences post \(\displaystyle g \circ h \) was given as the identity map on C and hence could be referred to as a bijection

Hi Bacterius,

Just as a further note, In drawing set theoretic maps of the situation of this problem, I am having some trouble finding a map g that works that is not injective as well as being surjective

Can you produce an example where g is not injective?

Peter

 

[Math Amateur]

Hi Bacterius,

Just as a further note, In drawing set theoretic maps of the situation of this problem, I am having some trouble finding a map g that works that is not injective as well as being surjective

Can you produce an example where g is not injective?

Peter

I am attaching a figure (see attached) that purports to show an example of the mappings g and h as discussed in the above posts where the following is the case:

(i) \(\displaystyle g \circ h = id_C \)

(ii) g is surjective, but not injective

(iii) h is injective but not surjective

Can someone confirm that my contentions regarding the attached figure are correct?

Mind you, this is not a proof that when \(\displaystyle g \circ h = id_C \) we have g is surjective and h is injective.

Can someone also help with the proof?

Peter

 

[Math Amateur]

I think I have a counterexample.

Let $B$ be a finite group with a subgroup $H$ such that $|G|=2$ and $|H|>1$. (Such groups exist).

Let $b\in B$ be an element of order $2$ in $B$. (Such a $b$ exists by Cauchy's theorem).

Define $C=\{1,-1\}$ under multiplication.

Define $g:B\to C$ as $g(h)=1$ for all $h\in H$ and $g( bh )=-1$ for all $h\in H$.

It can be seen that $g$ is a homomorphism.

Define $h:C\to B$ as $h(1)=e_B$ and $h(-1)=b$.

It's is easily seen that $h$ too is a homomorphism.

Note that $g(h(1))=g(e)=1$ and $g(h(-1))=g(b)=-1$.

Thus $g\circ h=id_C$. But neither $g$ is injective and nor $h$ is surjective.

Thanks caffeinemachine.

Indeed ... correct ... but I think I have now arrived at the view that g has to be surjective and h has to be injective ... which is the case in your example ...

... ... but you certainly demonstrated that the previous contention did not hold ... thanks ...

Peter

 

[Deveno]

Mea culpa, I got the order confused (as I sometimes do, I forget that $fg$ means "do $g$ first", and I think of $f$ being applied first).

Dumb, dumb, dumb!

 

[Math Amateur]

Mea culpa, I got the order confused (as I sometimes do, I forget that $fg$ means "do $g$ first", and I think of $f$ being applied first).

Dumb, dumb, dumb!

Well it got Bacterius and me thinking anyway ... and furthermore, your post was extremely informative and helpful ...

By the way, I was watching some online lectures on category theory by Steve Awodey and he was referring to \(\displaystyle f \circ g \) as "f after g" and I found that verbalization most helpful ...

Thanks again for your help,

Peter