Solving Simple Group Homomorphism Problem: Proving phi(G) is Subgroup of N

[A.Magnus]

I have this problem on simple group's homomorphism:

Let $G′$ be a group and let $\phi$ be a homomorphism from $G$ to $G′$. Assume that $G$ is simple, that $|G| \neq 2$, and that $G′$ has a normal subgroup $N$ of index 2. Show that $\phi (G) \subset N$.
​

And last year somebody gave me the following solution but which I am not quite understand, especially the last two parts:

(a) Let $\phi$ be homomorphism $\phi : G \to G'$ and $\sigma$ be the canonical homomorphis $\sigma : G′ \to G′/N$, such that

$G \to G' \to G'/N$.
​

(b) And let the composition of two mappings $\psi = \sigma \circ \phi$.
(c) Since $|G′/N|=2$ and $|G| \neq 2, \ \psi$ cannot be injective. So $ker (\psi)=G$.
(d) In other words, $\psi (G) = \{eN\} = N$.
(e) This implies $\sigma ( \phi (G)) = N$, which means $\phi (G) \subset N$.​

Since the responder may no longer be reached after so many months have passed, I am therefore posting questions here, hoping that some of you would help me with the last two parts:

(1) I understand up to (c), but how do you go from (c) to (d)?
(2) And also, how do you go from (d) to (e)?​

Thank you very much for your time and help.

 

[MostlyHarmless]

I'm not entirely familiar with all the definitions being used here but, I believe If $ker (\psi)=G$ then that says that $\psi$ is the trivial mapping, sending everything to the identity. So, I think $\psi(G)$ would then map to the identity in G', and then the identity in G' has to map to the trivial coset {eN}=N. d and e seem to me like one step, but again I could be wrong. I hope this helps.

 

[A.Magnus]

@MostlyHarmless : Thanks for your response! Actually, your analysis is pretty close to what I got after I first asked the question:

~~~~~~~~~~~~~~~~~~~~~~~~

(a) Let $\phi$ be homomorphism $\phi : G \to G'$ and $\sigma$ be the canonical homomorphis $\sigma : G′ \to G′/N$, and let the composition of two mappings $\psi = \sigma \cdot \phi$:

$G \underbrace{\overbrace{\to}^{\phi} G' \overbrace{\to}^{\sigma}}_{\psi = \sigma \circ \phi} G'/N$.
​

(b) Since $|G′/N|=2$ and $|G| \neq 2, \ \psi$ cannot be injective, meaning that $ker(\psi) \neq \{1\}$.
(c) Since $G$ is simple whose normal subgroup is either $G$ itself or $\{1\}$, and since $ker(\psi)$ is normal subgroup of $G$, therefore $ker(\psi)=G$.
(d) In other words, $\psi (G) = \{eN\}$, where $\{eN\}$ is the neutral element of $G'/N$, that is, $\psi (G) = \{eN\} = N$.
(e) This implies $\psi (G) = \sigma ( \phi (G)) = N$ , which implies $\phi (G) \subseteq N$.

~~~~~~~~~~~~~~~~~~~~~~~~

However, I am still puzzled by step (e), $\sigma ( \phi (G)) = N$ implies $\phi (G) \subseteq N$. Does this generalize into these: In $f: A \to B$, if $f(A) = B$, then this implies $A \subseteq B$?

Thanks again for your time and effort.

 

[MostlyHarmless]

I'm still thinking about e but I don't think that this generalizes to what you said. If you have f as you said above, f(A)=B just says that f maps A one to one and onto B. It doesn't for example let A={1, 2} B={3, 6} define f from A to B by f(1)=3 f(2)=6. So f(A)=B but A is not contained in B.

 

[A.Magnus]

I have also been musing about (e) with the same example as yours. I am very comfortable with the the logic of (a) to (d), but very doubtful about (e). I got these analysis from another forum. The responder has been helping generously me with lots of questions, I just don't feel comfortable confronting him/her with the example. Thanks again for your response!

 

[Dick]

I have also been musing about (e) with the same example as yours. I am very comfortable with the the logic of (a) to (d), but very doubtful about (e). I got these analysis from another forum. The responder has been helping generously me with lots of questions, I just don't feel comfortable confronting him/her with the example. Thanks again for your response!

I'm not sure I see what the problem is. Just think through what everything means. $G'/N$ has two cosets, call them $eN$ and $gN$. Every element of the group belongs to one of those two cosets. $\sigma$ just maps each element of $G'$ to the coset that contains it. $\sigma ( \phi (G)) = N$ must mean that every element of $\phi(G)$ must map to the coset $eN$, therefore every element of $\phi(G)$ belongs to the coset $eN$, therefore every element of $\phi(G)$ is contained in $N$.

 

[A.Magnus]

... therefore every element of $\phi(G)$ belongs to the coset $eN$, therefore every element of $\phi(G)$ is contained in $N$.

Thanks! But the statement you said above is the one that I don't understand, because it does not generalize. @MostlyHarmless also said the same thing when he/she wrote these:

... If you have $f$ as you said above, $f(A) = B$, [which] says that $f$ maps $A$ one to one and onto $B$. [Then] it doesn't [make sense, because] for example, let $A=\{1, 2\}, B=\{3, 6\}$ define $f$ from $A$ to $B$ by $f(1)=3, f(2)=6$. So $f(A)=B$ but $A$ is not contained in $B$.

Thanks again for your response.

 

[Dick]

Thanks! But the statement you said above is the one that I don't understand, because it does not generalize. @MostlyHarmless also said the same thing when he/she wrote these:
Thanks again for your response.

Something isn't false because you can make up a generalization that isn't true. Think about what the mapping $\sigma$ actually means.

 

[MostlyHarmless]

Just to be clear, I was simply showing that what you said was not true in general. What Dick said in his first post explains it pretty clearly to me.

 

[A.Magnus]

Something isn't false because you can make up a generalization that isn't true. Think about what the mapping σ\sigma actually means.

You got a point here, I will come back again to take another look. Thanks again.

 

[A.Magnus]

Here is what I managed to wrap up after getting helps from @Dick and @MostlyHarmless. Thanks to all your time and effort:

(1) Let $\phi$ be homomorphism $\phi:G \to G′$ and $\sigma$ be the canonical homomorphis $\sigma:G′ \to G′/N$, and let the composition of two mappings be $\psi = \sigma \cdot \phi$:

$G \overbrace{\underbrace{\to}_{\phi} G′\underbrace{\to}_{\sigma}}^{\psi} G′/N.$​

(2) Since $|G′/N|=2$ and $|G|\neq 2$, $\psi$ cannot be injective, meaning that $ker(\psi)\neq \{1\}$.
(3) Since $G$ is simple whose normal subgroup is either $G$ itself or $\{1\}$, and since $ker(\psi)$ is normal subgroup of $G$, therefore it has to be $ker(\psi)=G$.
(4) In other words, $\psi(G)={eN}$, where $e$ is the neutral element of $G'$ and ${eN}$ is the neutral element of $G′/N$, that is, $\psi(G) = \sigma (\phi (G)) = {eN}$.
(5) By definition of $\sigma$:

$\begin{align} \sigma : G' &\to G'/N \tag{5a}\\ \sigma : \phi(G) &\to \phi(G)/N, \quad \text{since} \ \phi(G) \subseteq (G) \tag{5b}\\ \sigma : \phi(g) &\mapsto \phi(g)N, \quad \forall g \in G \tag{5c}\\ \sigma (\phi(G)) &= \phi(G)N \tag{5d}\\ \end{align}$​

(6) By comparing the results of (4) and (5d), we conclude that $\phi(G) = \{e\}$.
(7) Since $\{e\} \subseteq N$ therefore $\phi(G) \subseteq N. \blacksquare$

 

[MostlyHarmless]

I'm not sure that $\phi(G)={e}$ Then you would have $\sigma$ mapping the identity to things that aren't the identity, which since $\sigma$ is a homomorphism, can't happen. In other words, if you have a function the identity to everything, then that function is not going to be well defined.

Certainly, it could be true that $\phi(G)={e}$, but only if $N={e}$

Also, eN isn't the identity element in G'/N. It is the set made of multiplying all the elements of N by e, the identity on the left. The identity in G'/N is e. It's obtained by multiplying by e in G' by e in N.

Dick's first post explained it well.

Basically, the function $\sigma$ takes elements of G' and says, if this element is of the form g*n then send it to the coset gN, and if it's of the form e*n send it to eN. One of these must be true since G' is a group and N is a subgroup of G', i.e. G' is closed. So since, $\phi(G)$ is a subset of G', $\phi(G)$ will satisfy one of the above conditions. So once $\sigma(\phi(G))=N$ is established that tells us that every element of $\phi(G)$ is of the form e*n, because of the above conditions.

 

[Dick]

$\sigma (\phi(G)) = \phi(G)N=eN$ does not tell you that $\phi(G)={e}$ it just tells you that $\phi(G)$ is a subset of $N$.

 

[A.Magnus]

σ(ϕ(G))=ϕ(G)N=eN does not tell you that ϕ(G)=e it just tells you that ϕ(G) is a subset of N.

Trust me, I have been working very hard trying to understand $\sigma (\phi(G)) = \phi(G)N=eN$ implies that $\phi (G) \subseteq N$, as I have been asking around http://www.quora.com/How-does-%CF%83-%CF%95-G-N-imply-%CF%95-G-%E2%8A%82N-1 and also here. Of course I could just scribble down the solution and turn in the paper next month, the professor won't have time to look at the detail anyway. However, being a HS math teacher myself, I don't think that is a right attitude -- I am curious about group theory. Here is my request:

Could you please indulge me by breaking down in step-by-step detail that $\sigma (\phi (G)) = \phi (G)N = eN$ implies that $\phi (G) \subseteq N$?

Thank you very much for your time and help.

 

[Dick]

Trust me, I have been working very hard trying to understand $\sigma (\phi(G)) = \phi(G)N=eN$ implies that $\phi (G) \subseteq N$, as I have been asking around http://www.quora.com/How-does-%CF%83-%CF%95-G-N-imply-%CF%95-G-%E2%8A%82N-1 and also here. Of course I could just scribble down the solution and turn in the paper next month, the professor won't have time to look at the detail anyway. However, being a HS math teacher myself, I don't think that is a right attitude -- I am curious about group theory. Here is my request:

Could you please indulge me by breaking down in step-by-step detail that $\sigma (\phi (G)) = \phi (G)N = eN$ implies that $\phi (G) \subseteq N$?

Thank you very much for your time and help.

Best if you take a crack at it yourself first. You want to show that $gN=N$ if and only if $g$ is an element of $N$. In other words for every element $n$ of $N$ you have to ask whether the equation $gm=n$ has a solution where $m$ is also an element of $N$. Just try it. It's not very hard. Just use that $N$ is a subgroup.

 

[A.Magnus]

Best if you take a crack at it yourself first. You want to show that gN=NgN=N if and only if gg is an element of NN. In other words for every element nn of NN you have to ask whether the equation gm=ngm=n has a solution where mm is also an element of NN. Just try it. It's not very hard. Just use that NN is a subgroup.

Thanks, I will take another look and report back to you shortly. Take care!

 

[A.Magnus]

Best if you take a crack at it yourself first. You want to show that $gN=N$ if and only if $g$ is an element of $N$. In other words for every element $n$ of $N$ you have to ask whether the equation $gm=n$ has a solution where $m$ is also an element of $N$. Just try it. It's not very hard. Just use that $N$ is a subgroup.

I think I get it now, hopefully I am get it squared out this time around, thanks to @MostlyHarmless also:

$\begin{align} \sigma : G' &\to G'/N \tag{1} \\ \phi (G) &\to \phi (G) /N \quad \text{since } \phi(G) \subseteq G' \tag{2} \\ g' &\mapsto \underbrace{g'N}_{= N} \quad \forall g' \in \phi(G) \tag{3} \\ \end{align}$

(4) From $g'N = N$, we are forced to conclude that $g' \in N$, for otherwise $g'N \neq N$
(5) The above line implies that $\forall g' \in \phi(G) \rightarrow g' \in N$, which further implies that $\phi(G) \subseteq N \qquad \blacksquare$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

One side question while you are here: Are you familiar with a question that I posted almost all over the places here and http://www.quora.com/How-do-I-prove-G-Z-R-G-is-isomorphic-to-Aut-R-G/answer/Charles-Slade?__snids__=919690641&__nsrc__=2 ? I got only lukewarm responses there but am still childishly curious about it. If you are, I will post it here. Thanks again.

 

[Dick]

(4) From $g'N = N$, we are forced to conclude that $g' \in N$, for otherwise $g'N \neq N$

This is the part that I was wondering if you could actually prove rather than just state. And, no, I'm not really familiar with the terminology in the other group problem.

 

[A.Magnus]

This is the part that I was wondering if you could actually prove rather than just state. And, no, I'm not really familiar with the terminology in the other group problem.

I think the proof is rather straightforward:

(1) Assume by contrary that $g' \notin N$
(2) Then $\phi (G)N = N$ and then $\phi (G) = NN^{-1} = \{e\}$
(3) But since $\phi$ is a hormomorphism, therefore $G = \{e\}$ too, which renders the whole mappings, $\sigma$ and also $\psi$, not well-defined.
(4) Therefore $g' \in N. \qquad \blacksquare$

I hope everything is cleared up now. Thanks again and again to all who have read this posting.

PS. As to your response that you are not familiar with my other question, I think that is quite understandable because this is a physics forum and I suspect first and foremost you are a physicist, and you know group theory as a by-product of being a physicist. Anyway I have posted the question here.

Thank you again and again to all.

 

[Dick]

I think the proof is rather straightforward:

(1) Assume by contrary that $g' \notin N$
(2) Then $\phi (G)N = N$ and then $\phi (G) = NN^{-1} = \{e\}$
(3) But since $\phi$ is a hormomorphism, therefore $G = \{e\}$ too, which renders the whole mappings, $\sigma$ and also $\psi$, not well-defined.
(4) Therefore $g' \in N. \qquad \blacksquare$

I hope everything is cleared up now. Thanks again and again to all who have read this posting.

PS. As to your response that you are not familiar with my other question, I think that is quite understandable because this is a physics forum and I suspect first and foremost you are a physicist, and you know group theory as a by-product of being a physicist. Anyway I have posted the question here.

Thank you again and again to all.

No, it's not cleared up. There's very little in that 'proof' that makes any sense. Since this is sort of basic to the notion of 'coset' I think it's worth figuring this out. Start by telling me what's wrong with saying $NN^{-1}=\{e\}$? What does it mean to write the product of two sets? And I don't see what $g'$ has to do with the whole middle of the proof.

 

[A.Magnus]

No, it's not cleared up. There's very little in that 'proof' that makes any sense. Since this is sort of basic to the notion of 'coset' I think it's worth figuring this out. Start by telling me what's wrong with saying $NN^{-1}=\{e\}$? What does it mean to write the product of two sets? And I don't see what $g'$ has to do with the whole middle of the proof.

I surely will take a look again. Thanks again!

 

[Dick]

I surely will take a look again. Thanks again!

Keep at it. You'll get this. But here's a start. If $g'N=N$ then for every $n$ in $N$ there must be an $m$ in $N$ such that $g'm=n$, right? Now just use that $m$ and $n$ are in the SUBGROUP $N$. No deep thinking required.

 

[A.Magnus]

Keep at it. You'll get this. But here's a start. If $g'N=N$ then for every $n$ in $N$ there must be an $m$ in $N$ such that $g'm=n$, right? Now just use that $m$ and $n$ are in the SUBGROUP $N$. No deep thinking required.

Can I say it this way instead: Since multiplying two subgroups is just like multiplying two polynomials, therefore

$\begin{align} \phi(G)N &= N \\ \phi(G) &= NN^{-1} \\ \phi(G) &\neq \{e\} \text{, but} \\ \phi(G) &= N \text{ instead.}\\ \end{align}$
​

But the next step is a little tricky. Recall that here I need to prove that $\phi(G) \subseteq N$. Can I just say that since $\phi(G) = N$, therefore $\forall x \in \phi(G) \rightarrow N$, thus $\phi(G) \subseteq N$? Let me know and thanks again for all your time and help.

 

[Dick]

Can I say it this way instead: Since multiplying two subgroups is just like multiplying two polynomials, therefore

$\begin{align} \phi(G)N &= N \\ \phi(G) &= NN^{-1} \\ \phi(G) &\neq \{e\} \text{, but} \\ \phi(G) &= N \text{ instead.}\\ \end{align}$
​

But the next step is a little tricky. Recall that here I need to prove that $\phi(G) \subseteq N$. Can I just say that since $\phi(G) = N$, therefore $\forall x \in \phi(G) \rightarrow N$, thus $\phi(G) \subseteq N$? Let me know and thanks again for all your time and help.

You really have to think harder about this. You can't do that. This is BASIC group theory. Suppose $N$ is the multiplicative group of nonzero integers mod 5. So $N=\{1,2,3,4\}$. Suppose $\phi(G)$ is the normal subgroup $\{1,4\}$. Convince yourself that the setup is right. Now figure out what's wrong with your argument. Multiplying subgroups is not like multiplying polynomials. $\phi(G)N=N$. But $\phi(G)$ is not equal to either $\{e\}$ or $N$. Go back to my post #22 and work from there.

 

[A.Magnus]

You really have to think harder about this. You can't do that. This is BASIC group theory. Suppose $N$ is the multiplicative group of nonzero integers mod 5. So $N=\{1,2,3,4\}$. Suppose $\phi(G)$ is the normal subgroup $\{1,4\}$. Convince yourself that the setup is right. Now figure out what's wrong with your argument. Multiplying subgroups is not like multiplying polynomials. $\phi(G)N=N$. But $\phi(G)$ is not equal to either $\{e\}$ or $N$. Go back to my post #22 and work from there.

I am not going to give up on this problem, but I am tending some other classes right now. I will get back here as soon as I am done with them. Thanks again and again for helping me on this basic question.

 

[Dick]

I am not going to give up on this problem, but I am tending some other classes right now. I will get back here as soon as I am done with them. Thanks again and again for helping me on this basic question.

Very welcome. And yes, don't give up on this one. There's some important points here.

 

[A.Magnus]

Very welcome. And yes, don't give up on this one. There's some important points here.

(1) Here is my latest thought, I trust I am right this time around since I did not have any bad gut feeling:

$\phi(G)N = N$ implies
$\forall g' \in \phi(G), \forall m \in N, \exists n \in N,$ such that $g'm = n$
$g'm = n \Rightarrow g' = nm^{-1}$
therefore $\forall g' \in \phi(G) \Rightarrow g' \in N$ since $g' = nm^{-1},$ and $n, m, m^{-1} \in N$
thus $\phi(G) \subseteq N.$
​

(2) Since you are here, I have a question pertaining to my last posting: Suppose that $A$ is a group, is it conventional to write $A^{-1}$, whose elements are the inverse of the elements of $A$? Or it applies only to certain group?

Thank you again and again for coaching me step by step to this very moment. I bid you a wonderful weekend tomorrow with your loved ones.

 

[Dick]

(1) Here is my latest thought, I trust I am right this time around since I did not have any bad gut feeling:

$\phi(G)N = N$ implies
$\forall g' \in \phi(G), \forall m \in N, \exists n \in N,$ such that $g'm = n$
$g'm = n \Rightarrow g' = nm^{-1}$
therefore $\forall g' \in \phi(G) \Rightarrow g' \in N$ since $g' = nm^{-1},$ and $n, m, m^{-1} \in N$
thus $\phi(G) \subseteq N.$
​

(2) Since you are here, I have a question pertaining to my last posting: Suppose that $A$ is a group, is it conventional to write $A^{-1}$, whose elements are the inverse of the elements of $A$? Or it applies only to certain group?

Thank you again and again for coaching me step by step to this very moment. I bid you a wonderful weekend tomorrow with your loved ones.

That's it and you're very welcome. If $A$ is a group then the inverse of every element is also in $A$. Remember these are sets, they don't have any order. So $A^{-1}$ is the same set as $A$. Isn't it?

 

[A.Magnus]

That's it and you're very welcome. If $A$ is a group then the inverse of every element is also in $A$. Remember these are sets, they don't have any order. So $A^{-1}$ is the same set as $A$. Isn't it?

Thanks to you first and foremost for tending this slowpoke guy, I am glad I met you. To answer your question: Yes, if $A = \mathbb Z_5^x$ just like the example you gave me, then $A^{-1} = \mathbb Z^x_5$ also. Thanks again.

 

[Dick]

Thanks to you first and foremost for tending this slowpoke guy, I am glad I met you. To answer your question: Yes, if $A = \mathbb Z_5^x$ just like the example you gave me, then $A^{-1} = \mathbb Z^x_5$ also. Thanks again.

If $A$ isn't a group then $A^{-1}$ might be a different set. If you want one more question to test your understanding, then if $A$ is a group, what is $AA$?

 

[A.Magnus]

If $A$ isn't a group then $A^{-1}$ might be a different set. If you want one more question to test your understanding, then if $A$ is a group, what is $AA$?

I think $AA = A^2 = A$, because all elements are closed by operation. Correct?

 

[Dick]

I think $AA = A^2 = A$, because all elements are closed by operation. Correct?

Correct. Another way to see it is that $e$ is an element of $A$. So $A=eA \subset AA$, and, of course $AA$ can't contain any elements that aren't in $A$ since group multiplication is a closed operation. As you say. I think you've got it.