Mixing
Homomorphism $f: G to H$ maps $e_g$ to $e_h$
Clash Royale CLAN TAG#URR8PPP
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I'm trying to wok out the proof that a homomorphism, $f: G to H$ maps the identity in one group to the identity in another. Several other answers provide an explanation for this, but there is one step I am confused on and haven't been able to find by browsing past answers. It's likely that I'm misunderstanding the justification for drawing on the inverse of $f$ and the logic which follows.
starting with the standard property of homorphisms, $f(g_1 * g_2) = f(g_1) oplus f(g_2)$, taking $*$ to be the binary operation in $G$, $oplus$ to the binary operation in $H$, and $g_1, g_2 in G$.
From here, the proof lets $g_1 = g_2 = e_g$, where $e_g$ is the identity of the group, $G$, which implies that $f(e_g * e_g) = f(e_g) * f(e_g)$, so $f(e_g) = f(e_g) * f(e_g)$.
Then, we multiply on the left by $left(f(e_g)right)^-1$, which should cause this to simplfiy to the identity on the left and $f(e_g)$ on the right. Here is my confusion:
(a) How do we conclude that such an inverse exists? I believe that a homorphism is by definition bijective and, by extension, invertible, as establishing such a function is necessary to establish that two groups are isomorphic. Is this the argument? Or is there an additional condition specific to the particular groups?
(b) Which binary operation are we applying when we multiply by this inverse? Clearly, $f(e_g) in H$ by the definition of $f$, which has the operation $oplus$, but $left(f(e_g)right)^-1 in G$, which has the operation, $*$. How can we 'mix' operations in this sense between elements of two different groups?
Thanks.
group-theory proof-explanation
asked Jul 15 at 0:55
Matt.P
768313
add a comment |Â
up vote
2
down vote
favorite
I'm trying to wok out the proof that a homomorphism, $f: G to H$ maps the identity in one group to the identity in another. Several other answers provide an explanation for this, but there is one step I am confused on and haven't been able to find by browsing past answers. It's likely that I'm misunderstanding the justification for drawing on the inverse of $f$ and the logic which follows.
starting with the standard property of homorphisms, $f(g_1 * g_2) = f(g_1) oplus f(g_2)$, taking $*$ to be the binary operation in $G$, $oplus$ to the binary operation in $H$, and $g_1, g_2 in G$.
From here, the proof lets $g_1 = g_2 = e_g$, where $e_g$ is the identity of the group, $G$, which implies that $f(e_g * e_g) = f(e_g) * f(e_g)$, so $f(e_g) = f(e_g) * f(e_g)$.
Then, we multiply on the left by $left(f(e_g)right)^-1$, which should cause this to simplfiy to the identity on the left and $f(e_g)$ on the right. Here is my confusion:
(a) How do we conclude that such an inverse exists? I believe that a homorphism is by definition bijective and, by extension, invertible, as establishing such a function is necessary to establish that two groups are isomorphic. Is this the argument? Or is there an additional condition specific to the particular groups?
(b) Which binary operation are we applying when we multiply by this inverse? Clearly, $f(e_g) in H$ by the definition of $f$, which has the operation $oplus$, but $left(f(e_g)right)^-1 in G$, which has the operation, $*$. How can we 'mix' operations in this sense between elements of two different groups?
Thanks.
group-theory proof-explanation
asked Jul 15 at 0:55
Matt.P
768313
The inverse exists because $H$ is a group. You're applying the operation of $H$.
– Randall
Jul 15 at 0:58
You also moved from calling the operation of $H$ $oplus$ to $*$ about halfway through, which created your confusion in question (b).
– Randall
Jul 15 at 0:59
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to wok out the proof that a homomorphism, $f: G to H$ maps the identity in one group to the identity in another. Several other answers provide an explanation for this, but there is one step I am confused on and haven't been able to find by browsing past answers. It's likely that I'm misunderstanding the justification for drawing on the inverse of $f$ and the logic which follows.
starting with the standard property of homorphisms, $f(g_1 * g_2) = f(g_1) oplus f(g_2)$, taking $*$ to be the binary operation in $G$, $oplus$ to the binary operation in $H$, and $g_1, g_2 in G$.
From here, the proof lets $g_1 = g_2 = e_g$, where $e_g$ is the identity of the group, $G$, which implies that $f(e_g * e_g) = f(e_g) * f(e_g)$, so $f(e_g) = f(e_g) * f(e_g)$.
Then, we multiply on the left by $left(f(e_g)right)^-1$, which should cause this to simplfiy to the identity on the left and $f(e_g)$ on the right. Here is my confusion:
(a) How do we conclude that such an inverse exists? I believe that a homorphism is by definition bijective and, by extension, invertible, as establishing such a function is necessary to establish that two groups are isomorphic. Is this the argument? Or is there an additional condition specific to the particular groups?
(b) Which binary operation are we applying when we multiply by this inverse? Clearly, $f(e_g) in H$ by the definition of $f$, which has the operation $oplus$, but $left(f(e_g)right)^-1 in G$, which has the operation, $*$. How can we 'mix' operations in this sense between elements of two different groups?
Thanks.
group-theory proof-explanation
asked Jul 15 at 0:55
Matt.P
768313
I'm trying to wok out the proof that a homomorphism, $f: G to H$ maps the identity in one group to the identity in another. Several other answers provide an explanation for this, but there is one step I am confused on and haven't been able to find by browsing past answers. It's likely that I'm misunderstanding the justification for drawing on the inverse of $f$ and the logic which follows.
starting with the standard property of homorphisms, $f(g_1 * g_2) = f(g_1) oplus f(g_2)$, taking $*$ to be the binary operation in $G$, $oplus$ to the binary operation in $H$, and $g_1, g_2 in G$.
From here, the proof lets $g_1 = g_2 = e_g$, where $e_g$ is the identity of the group, $G$, which implies that $f(e_g * e_g) = f(e_g) * f(e_g)$, so $f(e_g) = f(e_g) * f(e_g)$.
Then, we multiply on the left by $left(f(e_g)right)^-1$, which should cause this to simplfiy to the identity on the left and $f(e_g)$ on the right. Here is my confusion:
(a) How do we conclude that such an inverse exists? I believe that a homorphism is by definition bijective and, by extension, invertible, as establishing such a function is necessary to establish that two groups are isomorphic. Is this the argument? Or is there an additional condition specific to the particular groups?
(b) Which binary operation are we applying when we multiply by this inverse? Clearly, $f(e_g) in H$ by the definition of $f$, which has the operation $oplus$, but $left(f(e_g)right)^-1 in G$, which has the operation, $*$. How can we 'mix' operations in this sense between elements of two different groups?
Thanks.
group-theory proof-explanation
asked Jul 15 at 0:55
Matt.P
768313
asked Jul 15 at 0:55
Matt.P
768313
asked Jul 15 at 0:55
Matt.P
768313
asked Jul 15 at 0:55
Matt.P
768313
768313
The inverse exists because $H$ is a group. You're applying the operation of $H$.
– Randall
Jul 15 at 0:58
You also moved from calling the operation of $H$ $oplus$ to $*$ about halfway through, which created your confusion in question (b).
– Randall
Jul 15 at 0:59
add a comment |Â
The inverse exists because $H$ is a group. You're applying the operation of $H$.
– Randall
Jul 15 at 0:58
You also moved from calling the operation of $H$ $oplus$ to $*$ about halfway through, which created your confusion in question (b).
– Randall
Jul 15 at 0:59
The inverse exists because $H$ is a group. You're applying the operation of $H$.
– Randall
Jul 15 at 0:58
The inverse exists because $H$ is a group. You're applying the operation of $H$.
– Randall
Jul 15 at 0:58
You also moved from calling the operation of $H$ $oplus$ to $*$ about halfway through, which created your confusion in question (b).
– Randall
Jul 15 at 0:59
You also moved from calling the operation of $H$ $oplus$ to $*$ about halfway through, which created your confusion in question (b).
– Randall
Jul 15 at 0:59
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Here's what it should look like:
$$
f(e_G) = f(e_G * e_G) = f(e_G) oplus f(e_G).
$$
Ignore the middle term and write
$$
e_H oplus f(e_G) = f(e_G) oplus f(e_G).
$$
Note that all I have done is replace $f(e_G)$ by $e_H oplus f(e_G)$ which is legal by the identity property of $e_H$. Now operate on the right by $[f(e_G)]^-1$ computed in $H$.
answered Jul 15 at 1:01
Randall
7,2471825
Thanks for this. When you say 'operate on the right,' though, do you mean operate with respect to the group operation in H, in which case we write $[f(e_G)]^-1 oplus f(e_g)?$. Or do we use the operation from $G$, $*$? I think this is the only thing I'm still a bit confused on.
– Matt.P
Jul 15 at 1:16
1
My equation is in $H$, so everything is in $H$. "On the right" means $textblah blah blah oplus [f(e_G)]^-1$. Do that to both sides.
– Randall
Jul 15 at 1:22
add a comment |Â
up vote
2
down vote
How do we conclude that such an inverse exists?
Since $H$ is a group every element has its unique inverse.
I believe that a homorphism is by definition bijective
This is not true. Else terms like monomorphism, epimorphism or isomorphism would not make that much sense.
But I am sure you can give a counterexample pretty easily, if you try some homomorphisms.
ad b): You just use the property, that you have a homomorphism.
answered Jul 15 at 1:02
Cornman
2,53921128
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here's what it should look like:
$$
f(e_G) = f(e_G * e_G) = f(e_G) oplus f(e_G).
$$
Ignore the middle term and write
$$
e_H oplus f(e_G) = f(e_G) oplus f(e_G).
$$
Note that all I have done is replace $f(e_G)$ by $e_H oplus f(e_G)$ which is legal by the identity property of $e_H$. Now operate on the right by $[f(e_G)]^-1$ computed in $H$.
answered Jul 15 at 1:01
Randall
7,2471825
Thanks for this. When you say 'operate on the right,' though, do you mean operate with respect to the group operation in H, in which case we write $[f(e_G)]^-1 oplus f(e_g)?$. Or do we use the operation from $G$, $*$? I think this is the only thing I'm still a bit confused on.
– Matt.P
Jul 15 at 1:16
1
My equation is in $H$, so everything is in $H$. "On the right" means $textblah blah blah oplus [f(e_G)]^-1$. Do that to both sides.
– Randall
Jul 15 at 1:22
add a comment |Â
up vote
2
down vote
accepted
Here's what it should look like:
$$
f(e_G) = f(e_G * e_G) = f(e_G) oplus f(e_G).
$$
Ignore the middle term and write
$$
e_H oplus f(e_G) = f(e_G) oplus f(e_G).
$$
Note that all I have done is replace $f(e_G)$ by $e_H oplus f(e_G)$ which is legal by the identity property of $e_H$. Now operate on the right by $[f(e_G)]^-1$ computed in $H$.
answered Jul 15 at 1:01
Randall
7,2471825
Thanks for this. When you say 'operate on the right,' though, do you mean operate with respect to the group operation in H, in which case we write $[f(e_G)]^-1 oplus f(e_g)?$. Or do we use the operation from $G$, $*$? I think this is the only thing I'm still a bit confused on.
– Matt.P
Jul 15 at 1:16
1
My equation is in $H$, so everything is in $H$. "On the right" means $textblah blah blah oplus [f(e_G)]^-1$. Do that to both sides.
– Randall
Jul 15 at 1:22
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here's what it should look like:
$$
f(e_G) = f(e_G * e_G) = f(e_G) oplus f(e_G).
$$
Ignore the middle term and write
$$
e_H oplus f(e_G) = f(e_G) oplus f(e_G).
$$
Note that all I have done is replace $f(e_G)$ by $e_H oplus f(e_G)$ which is legal by the identity property of $e_H$. Now operate on the right by $[f(e_G)]^-1$ computed in $H$.
answered Jul 15 at 1:01
Randall
7,2471825
Here's what it should look like:
$$
f(e_G) = f(e_G * e_G) = f(e_G) oplus f(e_G).
$$
Ignore the middle term and write
$$
e_H oplus f(e_G) = f(e_G) oplus f(e_G).
$$
Note that all I have done is replace $f(e_G)$ by $e_H oplus f(e_G)$ which is legal by the identity property of $e_H$. Now operate on the right by $[f(e_G)]^-1$ computed in $H$.
answered Jul 15 at 1:01
Randall
7,2471825
answered Jul 15 at 1:01
Randall
7,2471825
answered Jul 15 at 1:01
Randall
7,2471825
answered Jul 15 at 1:01
Randall
7,2471825
7,2471825
Thanks for this. When you say 'operate on the right,' though, do you mean operate with respect to the group operation in H, in which case we write $[f(e_G)]^-1 oplus f(e_g)?$. Or do we use the operation from $G$, $*$? I think this is the only thing I'm still a bit confused on.
– Matt.P
Jul 15 at 1:16
1
My equation is in $H$, so everything is in $H$. "On the right" means $textblah blah blah oplus [f(e_G)]^-1$. Do that to both sides.
– Randall
Jul 15 at 1:22
add a comment |Â
Thanks for this. When you say 'operate on the right,' though, do you mean operate with respect to the group operation in H, in which case we write $[f(e_G)]^-1 oplus f(e_g)?$. Or do we use the operation from $G$, $*$? I think this is the only thing I'm still a bit confused on.
– Matt.P
Jul 15 at 1:16
1
My equation is in $H$, so everything is in $H$. "On the right" means $textblah blah blah oplus [f(e_G)]^-1$. Do that to both sides.
– Randall
Jul 15 at 1:22
Thanks for this. When you say 'operate on the right,' though, do you mean operate with respect to the group operation in H, in which case we write $[f(e_G)]^-1 oplus f(e_g)?$. Or do we use the operation from $G$, $*$? I think this is the only thing I'm still a bit confused on.
– Matt.P
Jul 15 at 1:16
Thanks for this. When you say 'operate on the right,' though, do you mean operate with respect to the group operation in H, in which case we write $[f(e_G)]^-1 oplus f(e_g)?$. Or do we use the operation from $G$, $*$? I think this is the only thing I'm still a bit confused on.
– Matt.P
Jul 15 at 1:16
1
1
My equation is in $H$, so everything is in $H$. "On the right" means $textblah blah blah oplus [f(e_G)]^-1$. Do that to both sides.
– Randall
Jul 15 at 1:22
My equation is in $H$, so everything is in $H$. "On the right" means $textblah blah blah oplus [f(e_G)]^-1$. Do that to both sides.
– Randall
Jul 15 at 1:22
add a comment |Â
up vote
2
down vote
How do we conclude that such an inverse exists?
Since $H$ is a group every element has its unique inverse.
I believe that a homorphism is by definition bijective
This is not true. Else terms like monomorphism, epimorphism or isomorphism would not make that much sense.
But I am sure you can give a counterexample pretty easily, if you try some homomorphisms.
ad b): You just use the property, that you have a homomorphism.
answered Jul 15 at 1:02
Cornman
2,53921128
add a comment |Â
up vote
2
down vote
How do we conclude that such an inverse exists?
Since $H$ is a group every element has its unique inverse.
I believe that a homorphism is by definition bijective
This is not true. Else terms like monomorphism, epimorphism or isomorphism would not make that much sense.
But I am sure you can give a counterexample pretty easily, if you try some homomorphisms.
ad b): You just use the property, that you have a homomorphism.
answered Jul 15 at 1:02
Cornman
2,53921128
add a comment |Â
up vote
2
down vote
up vote
2
down vote
How do we conclude that such an inverse exists?
Since $H$ is a group every element has its unique inverse.
I believe that a homorphism is by definition bijective
This is not true. Else terms like monomorphism, epimorphism or isomorphism would not make that much sense.
But I am sure you can give a counterexample pretty easily, if you try some homomorphisms.
ad b): You just use the property, that you have a homomorphism.
answered Jul 15 at 1:02
Cornman
2,53921128
How do we conclude that such an inverse exists?
Since $H$ is a group every element has its unique inverse.
I believe that a homorphism is by definition bijective
This is not true. Else terms like monomorphism, epimorphism or isomorphism would not make that much sense.
But I am sure you can give a counterexample pretty easily, if you try some homomorphisms.
ad b): You just use the property, that you have a homomorphism.
answered Jul 15 at 1:02
Cornman
2,53921128
answered Jul 15 at 1:02
Cornman
2,53921128
answered Jul 15 at 1:02
Cornman
2,53921128
answered Jul 15 at 1:02
Cornman
2,53921128
2,53921128
add a comment |Â
add a comment |Â
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The inverse exists because $H$ is a group. You're applying the operation of $H$.
– Randall
Jul 15 at 0:58
You also moved from calling the operation of $H$ $oplus$ to $*$ about halfway through, which created your confusion in question (b).
– Randall
Jul 15 at 0:59