Let $f: G to H$ be a group homomorphism with $|ker(f)|=2$. Show that every element of $H$ has no preimage or...
$begingroup$
Let $f: G to H$ be a group homomorphism such that the order of $ker(f)$ is $2$. Show that any element of $H$ has either no pre-image or exactly two pre-images under $f$.
I have been trying to prove this -
since $|ker f|= 2$, $ker f$ contains two elements say $k_1, k_2$.
Now we need to show each element of $H$ contains exactly $2$ or no pre-images.
Suppose $y in H$ such that $y$ has $3$ pre-images, say $x_1,x_2,x_3$. Then $f(x_1)=f(x_2)=f(x_3)$.
Since $f$ is a homomorphism, $x₁cdot x_2^{−1},x_2cdot x_3^{-1},x_3cdot x_1^{-1}$ belong to $ker f$.
Without loss of generality suppose $x_2cdot x_3^{-1} = x_1cdot x_2^{−1} = k_1$, and this implies $x_1=x_2$.
Is this ok?
group-theory proof-verification group-homomorphism
edited Dec 22 '18 at 11:08
Shaun
9,814113684
asked Aug 1 '16 at 12:14
shin.jinishin.jini
112
$endgroup$
add a comment |
$begingroup$
Let $f: G to H$ be a group homomorphism such that the order of $ker(f)$ is $2$. Show that any element of $H$ has either no pre-image or exactly two pre-images under $f$.
I have been trying to prove this -
since $|ker f|= 2$, $ker f$ contains two elements say $k_1, k_2$.
Now we need to show each element of $H$ contains exactly $2$ or no pre-images.
Suppose $y in H$ such that $y$ has $3$ pre-images, say $x_1,x_2,x_3$. Then $f(x_1)=f(x_2)=f(x_3)$.
Since $f$ is a homomorphism, $x₁cdot x_2^{−1},x_2cdot x_3^{-1},x_3cdot x_1^{-1}$ belong to $ker f$.
Without loss of generality suppose $x_2cdot x_3^{-1} = x_1cdot x_2^{−1} = k_1$, and this implies $x_1=x_2$.
Is this ok?
group-theory proof-verification group-homomorphism
edited Dec 22 '18 at 11:08
Shaun
9,814113684
asked Aug 1 '16 at 12:14
shin.jinishin.jini
112
$endgroup$
$begingroup$
Welcome to Math SX! What is the question?
$endgroup$
– Bernard
Aug 1 '16 at 12:17
3
$begingroup$
Welcome to the site, shin.jini! Could you tell us how you have tried to solve this problem? Do you understand what the problem asks you to do?
$endgroup$
– Daan Michiels
Aug 1 '16 at 12:24
$begingroup$
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
$endgroup$
– Martin Sleziak
Aug 1 '16 at 17:24
add a comment |
$begingroup$
Let $f: G to H$ be a group homomorphism such that the order of $ker(f)$ is $2$. Show that any element of $H$ has either no pre-image or exactly two pre-images under $f$.
I have been trying to prove this -
since $|ker f|= 2$, $ker f$ contains two elements say $k_1, k_2$.
Now we need to show each element of $H$ contains exactly $2$ or no pre-images.
Suppose $y in H$ such that $y$ has $3$ pre-images, say $x_1,x_2,x_3$. Then $f(x_1)=f(x_2)=f(x_3)$.
Since $f$ is a homomorphism, $x₁cdot x_2^{−1},x_2cdot x_3^{-1},x_3cdot x_1^{-1}$ belong to $ker f$.
Without loss of generality suppose $x_2cdot x_3^{-1} = x_1cdot x_2^{−1} = k_1$, and this implies $x_1=x_2$.
Is this ok?
group-theory proof-verification group-homomorphism
edited Dec 22 '18 at 11:08
Shaun
9,814113684
asked Aug 1 '16 at 12:14
shin.jinishin.jini
112
$endgroup$
Let $f: G to H$ be a group homomorphism such that the order of $ker(f)$ is $2$. Show that any element of $H$ has either no pre-image or exactly two pre-images under $f$.
I have been trying to prove this -
since $|ker f|= 2$, $ker f$ contains two elements say $k_1, k_2$.
Now we need to show each element of $H$ contains exactly $2$ or no pre-images.
Suppose $y in H$ such that $y$ has $3$ pre-images, say $x_1,x_2,x_3$. Then $f(x_1)=f(x_2)=f(x_3)$.
Since $f$ is a homomorphism, $x₁cdot x_2^{−1},x_2cdot x_3^{-1},x_3cdot x_1^{-1}$ belong to $ker f$.
Without loss of generality suppose $x_2cdot x_3^{-1} = x_1cdot x_2^{−1} = k_1$, and this implies $x_1=x_2$.
Is this ok?
group-theory proof-verification group-homomorphism
group-theory proof-verification group-homomorphism
edited Dec 22 '18 at 11:08
Shaun
9,814113684
asked Aug 1 '16 at 12:14
shin.jinishin.jini
112
edited Dec 22 '18 at 11:08
Shaun
9,814113684
asked Aug 1 '16 at 12:14
shin.jinishin.jini
112
edited Dec 22 '18 at 11:08
Shaun
9,814113684
edited Dec 22 '18 at 11:08
Shaun
9,814113684
edited Dec 22 '18 at 11:08
Shaun
9,814113684
9,814113684
asked Aug 1 '16 at 12:14
shin.jinishin.jini
112
asked Aug 1 '16 at 12:14
shin.jinishin.jini
112
asked Aug 1 '16 at 12:14
shin.jinishin.jini
112
112
$begingroup$
Welcome to Math SX! What is the question?
$endgroup$
– Bernard
Aug 1 '16 at 12:17
3
$begingroup$
Welcome to the site, shin.jini! Could you tell us how you have tried to solve this problem? Do you understand what the problem asks you to do?
$endgroup$
– Daan Michiels
Aug 1 '16 at 12:24
$begingroup$
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
$endgroup$
– Martin Sleziak
Aug 1 '16 at 17:24
add a comment |
$begingroup$
Welcome to Math SX! What is the question?
$endgroup$
– Bernard
Aug 1 '16 at 12:17
3
$begingroup$
Welcome to the site, shin.jini! Could you tell us how you have tried to solve this problem? Do you understand what the problem asks you to do?
$endgroup$
– Daan Michiels
Aug 1 '16 at 12:24
$begingroup$
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
$endgroup$
– Martin Sleziak
Aug 1 '16 at 17:24
$begingroup$
Welcome to Math SX! What is the question?
$endgroup$
– Bernard
Aug 1 '16 at 12:17
$begingroup$
Welcome to Math SX! What is the question?
$endgroup$
– Bernard
Aug 1 '16 at 12:17
3
3
$begingroup$
Welcome to the site, shin.jini! Could you tell us how you have tried to solve this problem? Do you understand what the problem asks you to do?
$endgroup$
– Daan Michiels
Aug 1 '16 at 12:24
$begingroup$
Welcome to the site, shin.jini! Could you tell us how you have tried to solve this problem? Do you understand what the problem asks you to do?
$endgroup$
– Daan Michiels
Aug 1 '16 at 12:24
$begingroup$
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
$endgroup$
– Martin Sleziak
Aug 1 '16 at 17:24
$begingroup$
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
$endgroup$
– Martin Sleziak
Aug 1 '16 at 17:24
add a comment |
1 Answer
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That's a good try.
I take it that you're using the fact that proving $Alor B$ is the same as proving $(lnot A)to B$. That's fine.
One problem is that you haven't justified why, if at all, we can assume w.l.o.g. that $x_2x_3^{-1}=x_1 x_2^{-1}$. I'm not sure whether $x_1=x_2$ follows from this either. Even if it did, so would $x_2=x_3$, so you wouldn't have shown there is two pre-images.
Here's how to prove the result:
First of all, consider $e_H$. Since $f$ is a homomorphism, we have $f(e_G)=f(e_Ge_G)=f(e_G)f(e_G)in H$, which gives $f(e_G)=e_H$. Thus $e_H$ has a pre-image in $G$; indeed, since $lvertker frvert=2$ and $K=ker f={ kin Gmid f(k)=e_H}$, the existence of a pre-image is guaranteed and we know that there is exactly two of them.
Now let $hin H$. If $h$ has a pre-image, then $h=f(g)$ for some $gin G$. Define the set of pre-images of $h$ under $f$ as $f^{-1}(h):={g'in Gmid h=f(g')}$. Recall that $gK={gkin Gmid kin K}$. Now, by definition, we have that
$$f^{-1}(h)=gK,$$
but the map
$$begin{align}
I(g): K&to gK \
k&mapsto gk
end{align}$$
is clearly a bijection. Hence $f^{-1}(g)$ has the same number of elements as $K$. Thus $g$ has two pre-images.
answered Dec 22 '18 at 11:03
ShaunShaun
9,814113684
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
That's a good try.
I take it that you're using the fact that proving $Alor B$ is the same as proving $(lnot A)to B$. That's fine.
One problem is that you haven't justified why, if at all, we can assume w.l.o.g. that $x_2x_3^{-1}=x_1 x_2^{-1}$. I'm not sure whether $x_1=x_2$ follows from this either. Even if it did, so would $x_2=x_3$, so you wouldn't have shown there is two pre-images.
Here's how to prove the result:
First of all, consider $e_H$. Since $f$ is a homomorphism, we have $f(e_G)=f(e_Ge_G)=f(e_G)f(e_G)in H$, which gives $f(e_G)=e_H$. Thus $e_H$ has a pre-image in $G$; indeed, since $lvertker frvert=2$ and $K=ker f={ kin Gmid f(k)=e_H}$, the existence of a pre-image is guaranteed and we know that there is exactly two of them.
Now let $hin H$. If $h$ has a pre-image, then $h=f(g)$ for some $gin G$. Define the set of pre-images of $h$ under $f$ as $f^{-1}(h):={g'in Gmid h=f(g')}$. Recall that $gK={gkin Gmid kin K}$. Now, by definition, we have that
$$f^{-1}(h)=gK,$$
but the map
$$begin{align}
I(g): K&to gK \
k&mapsto gk
end{align}$$
is clearly a bijection. Hence $f^{-1}(g)$ has the same number of elements as $K$. Thus $g$ has two pre-images.
answered Dec 22 '18 at 11:03
ShaunShaun
9,814113684
$endgroup$
add a comment |
$begingroup$
That's a good try.
I take it that you're using the fact that proving $Alor B$ is the same as proving $(lnot A)to B$. That's fine.
One problem is that you haven't justified why, if at all, we can assume w.l.o.g. that $x_2x_3^{-1}=x_1 x_2^{-1}$. I'm not sure whether $x_1=x_2$ follows from this either. Even if it did, so would $x_2=x_3$, so you wouldn't have shown there is two pre-images.
Here's how to prove the result:
First of all, consider $e_H$. Since $f$ is a homomorphism, we have $f(e_G)=f(e_Ge_G)=f(e_G)f(e_G)in H$, which gives $f(e_G)=e_H$. Thus $e_H$ has a pre-image in $G$; indeed, since $lvertker frvert=2$ and $K=ker f={ kin Gmid f(k)=e_H}$, the existence of a pre-image is guaranteed and we know that there is exactly two of them.
Now let $hin H$. If $h$ has a pre-image, then $h=f(g)$ for some $gin G$. Define the set of pre-images of $h$ under $f$ as $f^{-1}(h):={g'in Gmid h=f(g')}$. Recall that $gK={gkin Gmid kin K}$. Now, by definition, we have that
$$f^{-1}(h)=gK,$$
but the map
$$begin{align}
I(g): K&to gK \
k&mapsto gk
end{align}$$
is clearly a bijection. Hence $f^{-1}(g)$ has the same number of elements as $K$. Thus $g$ has two pre-images.
answered Dec 22 '18 at 11:03
ShaunShaun
9,814113684
$endgroup$
add a comment |
$begingroup$
That's a good try.
I take it that you're using the fact that proving $Alor B$ is the same as proving $(lnot A)to B$. That's fine.
One problem is that you haven't justified why, if at all, we can assume w.l.o.g. that $x_2x_3^{-1}=x_1 x_2^{-1}$. I'm not sure whether $x_1=x_2$ follows from this either. Even if it did, so would $x_2=x_3$, so you wouldn't have shown there is two pre-images.
Here's how to prove the result:
First of all, consider $e_H$. Since $f$ is a homomorphism, we have $f(e_G)=f(e_Ge_G)=f(e_G)f(e_G)in H$, which gives $f(e_G)=e_H$. Thus $e_H$ has a pre-image in $G$; indeed, since $lvertker frvert=2$ and $K=ker f={ kin Gmid f(k)=e_H}$, the existence of a pre-image is guaranteed and we know that there is exactly two of them.
Now let $hin H$. If $h$ has a pre-image, then $h=f(g)$ for some $gin G$. Define the set of pre-images of $h$ under $f$ as $f^{-1}(h):={g'in Gmid h=f(g')}$. Recall that $gK={gkin Gmid kin K}$. Now, by definition, we have that
$$f^{-1}(h)=gK,$$
but the map
$$begin{align}
I(g): K&to gK \
k&mapsto gk
end{align}$$
is clearly a bijection. Hence $f^{-1}(g)$ has the same number of elements as $K$. Thus $g$ has two pre-images.
answered Dec 22 '18 at 11:03
ShaunShaun
9,814113684
$endgroup$
That's a good try.
I take it that you're using the fact that proving $Alor B$ is the same as proving $(lnot A)to B$. That's fine.
One problem is that you haven't justified why, if at all, we can assume w.l.o.g. that $x_2x_3^{-1}=x_1 x_2^{-1}$. I'm not sure whether $x_1=x_2$ follows from this either. Even if it did, so would $x_2=x_3$, so you wouldn't have shown there is two pre-images.
Here's how to prove the result:
First of all, consider $e_H$. Since $f$ is a homomorphism, we have $f(e_G)=f(e_Ge_G)=f(e_G)f(e_G)in H$, which gives $f(e_G)=e_H$. Thus $e_H$ has a pre-image in $G$; indeed, since $lvertker frvert=2$ and $K=ker f={ kin Gmid f(k)=e_H}$, the existence of a pre-image is guaranteed and we know that there is exactly two of them.
Now let $hin H$. If $h$ has a pre-image, then $h=f(g)$ for some $gin G$. Define the set of pre-images of $h$ under $f$ as $f^{-1}(h):={g'in Gmid h=f(g')}$. Recall that $gK={gkin Gmid kin K}$. Now, by definition, we have that
$$f^{-1}(h)=gK,$$
but the map
$$begin{align}
I(g): K&to gK \
k&mapsto gk
end{align}$$
is clearly a bijection. Hence $f^{-1}(g)$ has the same number of elements as $K$. Thus $g$ has two pre-images.
answered Dec 22 '18 at 11:03
ShaunShaun
9,814113684
edited Dec 22 '18 at 11:10
answered Dec 22 '18 at 11:03
ShaunShaun
9,814113684
answered Dec 22 '18 at 11:03
ShaunShaun
9,814113684
answered Dec 22 '18 at 11:03
ShaunShaun
9,814113684
9,814113684
add a comment |
add a comment |
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$begingroup$
Welcome to Math SX! What is the question?
$endgroup$
– Bernard
Aug 1 '16 at 12:17
3
$begingroup$
Welcome to the site, shin.jini! Could you tell us how you have tried to solve this problem? Do you understand what the problem asks you to do?
$endgroup$
– Daan Michiels
Aug 1 '16 at 12:24
$begingroup$
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
$endgroup$
– Martin Sleziak
Aug 1 '16 at 17:24