Mixing
Homomorphism from divisible group to finite group is always trivial
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Let $A$ be a divisible group, let $B$ be a finite group, and let $f: A rightarrow B$ be a homomorphism. Show that $f$ is trivial.
(A group $A$ is divisible if for each $a in A$ and $n ge 1$ there exists some $b in A$ such that $b^n = a$)
I wanted to know if my solution is correct -
Let $a in A$. And assume that $|B| = n$ for some $n in mathbbN$. So we can see that -
$f(a)=f(b^n)=(f(b))^n=e_B$ and therefore $f$ is trivial.
The first $=$ is because of $A$ being a divisible group, and the second is because of $f$ being an homomorphism.
group-theory finite-groups group-homomorphism divisible-groups
asked Jul 20 at 21:45
ChikChak
617216
add a comment |Â
up vote
4
down vote
favorite
Let $A$ be a divisible group, let $B$ be a finite group, and let $f: A rightarrow B$ be a homomorphism. Show that $f$ is trivial.
(A group $A$ is divisible if for each $a in A$ and $n ge 1$ there exists some $b in A$ such that $b^n = a$)
I wanted to know if my solution is correct -
Let $a in A$. And assume that $|B| = n$ for some $n in mathbbN$. So we can see that -
$f(a)=f(b^n)=(f(b))^n=e_B$ and therefore $f$ is trivial.
The first $=$ is because of $A$ being a divisible group, and the second is because of $f$ being an homomorphism.
group-theory finite-groups group-homomorphism divisible-groups
asked Jul 20 at 21:45
ChikChak
617216
1
Looks very good.
– Lee Mosher
Jul 20 at 21:51
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $A$ be a divisible group, let $B$ be a finite group, and let $f: A rightarrow B$ be a homomorphism. Show that $f$ is trivial.
(A group $A$ is divisible if for each $a in A$ and $n ge 1$ there exists some $b in A$ such that $b^n = a$)
I wanted to know if my solution is correct -
Let $a in A$. And assume that $|B| = n$ for some $n in mathbbN$. So we can see that -
$f(a)=f(b^n)=(f(b))^n=e_B$ and therefore $f$ is trivial.
The first $=$ is because of $A$ being a divisible group, and the second is because of $f$ being an homomorphism.
group-theory finite-groups group-homomorphism divisible-groups
asked Jul 20 at 21:45
ChikChak
617216
Let $A$ be a divisible group, let $B$ be a finite group, and let $f: A rightarrow B$ be a homomorphism. Show that $f$ is trivial.
(A group $A$ is divisible if for each $a in A$ and $n ge 1$ there exists some $b in A$ such that $b^n = a$)
I wanted to know if my solution is correct -
Let $a in A$. And assume that $|B| = n$ for some $n in mathbbN$. So we can see that -
$f(a)=f(b^n)=(f(b))^n=e_B$ and therefore $f$ is trivial.
The first $=$ is because of $A$ being a divisible group, and the second is because of $f$ being an homomorphism.
group-theory finite-groups group-homomorphism divisible-groups
asked Jul 20 at 21:45
ChikChak
617216
edited Jul 20 at 21:46
asked Jul 20 at 21:45
ChikChak
617216
asked Jul 20 at 21:45
ChikChak
617216
asked Jul 20 at 21:45
ChikChak
617216
617216
1
Looks very good.
– Lee Mosher
Jul 20 at 21:51
add a comment |Â
1
Looks very good.
– Lee Mosher
Jul 20 at 21:51
1
1
Looks very good.
– Lee Mosher
Jul 20 at 21:51
Looks very good.
– Lee Mosher
Jul 20 at 21:51
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The proof is correct, but I'd add something for clarity.
Suppose $A$ is divisible and that $B$ is finite with $|B|=n$. If $fcolon Ato B$ is a homomorphism and $ain A$, then there exists $xin A$ such that $x^n=a$. Therefore
$$
f(a)=f(x^n)=f(x)^n=e_B
$$
Since $a$ was arbitrary, we conclude that $f$ is trivial.
A similar technique shows that the homomorphic image of a divisible group is divisible. You may want to show that a nontrivial divisible group is infinite.
answered Jul 20 at 21:51
egreg
164k1180187
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The proof is correct, but I'd add something for clarity.
Suppose $A$ is divisible and that $B$ is finite with $|B|=n$. If $fcolon Ato B$ is a homomorphism and $ain A$, then there exists $xin A$ such that $x^n=a$. Therefore
$$
f(a)=f(x^n)=f(x)^n=e_B
$$
Since $a$ was arbitrary, we conclude that $f$ is trivial.
A similar technique shows that the homomorphic image of a divisible group is divisible. You may want to show that a nontrivial divisible group is infinite.
answered Jul 20 at 21:51
egreg
164k1180187
add a comment |Â
up vote
2
down vote
accepted
The proof is correct, but I'd add something for clarity.
Suppose $A$ is divisible and that $B$ is finite with $|B|=n$. If $fcolon Ato B$ is a homomorphism and $ain A$, then there exists $xin A$ such that $x^n=a$. Therefore
$$
f(a)=f(x^n)=f(x)^n=e_B
$$
Since $a$ was arbitrary, we conclude that $f$ is trivial.
A similar technique shows that the homomorphic image of a divisible group is divisible. You may want to show that a nontrivial divisible group is infinite.
answered Jul 20 at 21:51
egreg
164k1180187
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The proof is correct, but I'd add something for clarity.
Suppose $A$ is divisible and that $B$ is finite with $|B|=n$. If $fcolon Ato B$ is a homomorphism and $ain A$, then there exists $xin A$ such that $x^n=a$. Therefore
$$
f(a)=f(x^n)=f(x)^n=e_B
$$
Since $a$ was arbitrary, we conclude that $f$ is trivial.
A similar technique shows that the homomorphic image of a divisible group is divisible. You may want to show that a nontrivial divisible group is infinite.
answered Jul 20 at 21:51
egreg
164k1180187
The proof is correct, but I'd add something for clarity.
Suppose $A$ is divisible and that $B$ is finite with $|B|=n$. If $fcolon Ato B$ is a homomorphism and $ain A$, then there exists $xin A$ such that $x^n=a$. Therefore
$$
f(a)=f(x^n)=f(x)^n=e_B
$$
Since $a$ was arbitrary, we conclude that $f$ is trivial.
A similar technique shows that the homomorphic image of a divisible group is divisible. You may want to show that a nontrivial divisible group is infinite.
answered Jul 20 at 21:51
egreg
164k1180187
answered Jul 20 at 21:51
egreg
164k1180187
answered Jul 20 at 21:51
egreg
164k1180187
answered Jul 20 at 21:51
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858046%2fhomomorphism-from-divisible-group-to-finite-group-is-always-trivial%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Looks very good.
– Lee Mosher
Jul 20 at 21:51