Mixing
Suppose $a, bin G$ such that $lvert arvert$ is odd and $aba^{-1}=b^{-1}$. Show that $b^2=e$.
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.30.
Suppose $a$ and $b$ belong to a group, $a$ has odd order, and $aba^{-1}=b^{-1}$. Show that $b^2=e$.
My Attempt:
Let $lvert arvert=2n+1$, $ninBbb Ncup{0}$.
We have $$begin{align}
b&=(b^{-1})^{-1} \
&=(aba^{-1})^{-1} \
&=ab^{-1}a^{-1},
end{align}$$
so, by induction, $b=a(underbrace{aba^{-1}}_{=b^{-1}})a^{-1}=dots =a^{2n}ba^{-2n}$, so that
$$begin{align}
b&=a^{2n}ba^{-2n} \
&=a^{2n}(ab^{-1}a^{-1})a^{-2n} \
&=a^{2n+1}b^{-1}a^{-(2n+1)} \
&=b^{-1},
end{align}$$
so $b=b^{-1}$, i.e., $b^2=e$.$square$
Is the above proof okay? I think it is.
However, I'd be interested to see how properties of Baumslag-Solitar groups could be used here, since $aba^{-1}=b^{-1}$ brings them to mind.
group-theory proof-verification alternative-proof
asked Jan 14 at 16:33
ShaunShaun
9,221113684
$endgroup$
add a comment |
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.30.
Suppose $a$ and $b$ belong to a group, $a$ has odd order, and $aba^{-1}=b^{-1}$. Show that $b^2=e$.
My Attempt:
Let $lvert arvert=2n+1$, $ninBbb Ncup{0}$.
We have $$begin{align}
b&=(b^{-1})^{-1} \
&=(aba^{-1})^{-1} \
&=ab^{-1}a^{-1},
end{align}$$
so, by induction, $b=a(underbrace{aba^{-1}}_{=b^{-1}})a^{-1}=dots =a^{2n}ba^{-2n}$, so that
$$begin{align}
b&=a^{2n}ba^{-2n} \
&=a^{2n}(ab^{-1}a^{-1})a^{-2n} \
&=a^{2n+1}b^{-1}a^{-(2n+1)} \
&=b^{-1},
end{align}$$
so $b=b^{-1}$, i.e., $b^2=e$.$square$
Is the above proof okay? I think it is.
However, I'd be interested to see how properties of Baumslag-Solitar groups could be used here, since $aba^{-1}=b^{-1}$ brings them to mind.
group-theory proof-verification alternative-proof
asked Jan 14 at 16:33
ShaunShaun
9,221113684
$endgroup$
3
$begingroup$
Yes the proof is OK.
$endgroup$
– Derek Holt
Jan 14 at 16:37
2
$begingroup$
Yes it is perfect
$endgroup$
– greedoid
Jan 14 at 16:38
add a comment |
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.30.
Suppose $a$ and $b$ belong to a group, $a$ has odd order, and $aba^{-1}=b^{-1}$. Show that $b^2=e$.
My Attempt:
Let $lvert arvert=2n+1$, $ninBbb Ncup{0}$.
We have $$begin{align}
b&=(b^{-1})^{-1} \
&=(aba^{-1})^{-1} \
&=ab^{-1}a^{-1},
end{align}$$
so, by induction, $b=a(underbrace{aba^{-1}}_{=b^{-1}})a^{-1}=dots =a^{2n}ba^{-2n}$, so that
$$begin{align}
b&=a^{2n}ba^{-2n} \
&=a^{2n}(ab^{-1}a^{-1})a^{-2n} \
&=a^{2n+1}b^{-1}a^{-(2n+1)} \
&=b^{-1},
end{align}$$
so $b=b^{-1}$, i.e., $b^2=e$.$square$
Is the above proof okay? I think it is.
However, I'd be interested to see how properties of Baumslag-Solitar groups could be used here, since $aba^{-1}=b^{-1}$ brings them to mind.
group-theory proof-verification alternative-proof
asked Jan 14 at 16:33
ShaunShaun
9,221113684
$endgroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.30.
Suppose $a$ and $b$ belong to a group, $a$ has odd order, and $aba^{-1}=b^{-1}$. Show that $b^2=e$.
My Attempt:
Let $lvert arvert=2n+1$, $ninBbb Ncup{0}$.
We have $$begin{align}
b&=(b^{-1})^{-1} \
&=(aba^{-1})^{-1} \
&=ab^{-1}a^{-1},
end{align}$$
so, by induction, $b=a(underbrace{aba^{-1}}_{=b^{-1}})a^{-1}=dots =a^{2n}ba^{-2n}$, so that
$$begin{align}
b&=a^{2n}ba^{-2n} \
&=a^{2n}(ab^{-1}a^{-1})a^{-2n} \
&=a^{2n+1}b^{-1}a^{-(2n+1)} \
&=b^{-1},
end{align}$$
so $b=b^{-1}$, i.e., $b^2=e$.$square$
Is the above proof okay? I think it is.
However, I'd be interested to see how properties of Baumslag-Solitar groups could be used here, since $aba^{-1}=b^{-1}$ brings them to mind.
group-theory proof-verification alternative-proof
group-theory proof-verification alternative-proof
asked Jan 14 at 16:33
ShaunShaun
9,221113684
asked Jan 14 at 16:33
ShaunShaun
9,221113684
edited Jan 14 at 17:00
Shaun
asked Jan 14 at 16:33
ShaunShaun
9,221113684
asked Jan 14 at 16:33
ShaunShaun
9,221113684
asked Jan 14 at 16:33
ShaunShaun
9,221113684
9,221113684
3
$begingroup$
Yes the proof is OK.
$endgroup$
– Derek Holt
Jan 14 at 16:37
2
$begingroup$
Yes it is perfect
$endgroup$
– greedoid
Jan 14 at 16:38
add a comment |
3
$begingroup$
Yes the proof is OK.
$endgroup$
– Derek Holt
Jan 14 at 16:37
2
$begingroup$
Yes it is perfect
$endgroup$
– greedoid
Jan 14 at 16:38
3
3
$begingroup$
Yes the proof is OK.
$endgroup$
– Derek Holt
Jan 14 at 16:37
$begingroup$
Yes the proof is OK.
$endgroup$
– Derek Holt
Jan 14 at 16:37
2
2
$begingroup$
Yes it is perfect
$endgroup$
– greedoid
Jan 14 at 16:38
$begingroup$
Yes it is perfect
$endgroup$
– greedoid
Jan 14 at 16:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your proof is fine. Here is an alternative one using a bit more group theory.
Let $G$ denote the subgroup generated by $a$ and $b$. Let $H cong mathbb{Z}_n$ be the subgroup of $G$ generated by $b$. Since $aba^{-1}=b^{-1}$, it follows that $H$ is normal $G$. The quotient $G/H$ must be isomorphic to $mathbb{Z}_m$ for some odd $m$, since the image of $a$ under the map $G rightarrow G/H$ maps to a generator of $G/H$ and $a$ has odd order. The conjugation action of $a$ on $H$ hence gives rise to a homomorphism
$$ mathbb{Z}_m rightarrow mathrm{Aut}(H): 1 mapsto varphi=( b mapsto aba^{-1}=b^{-1}) $$
Note that $varphi$ is either trivial or of order two. Since $mvarphi =0$ for $m$ odd, it follows that $varphi$ is trivial. Hence $b=b^{-1}$, i.e. $b^2=e$.
edited Jan 14 at 19:21
Shaun
9,221113684
answered Jan 14 at 19:20
MortenMorten
605
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Your proof is fine. Here is an alternative one using a bit more group theory.
Let $G$ denote the subgroup generated by $a$ and $b$. Let $H cong mathbb{Z}_n$ be the subgroup of $G$ generated by $b$. Since $aba^{-1}=b^{-1}$, it follows that $H$ is normal $G$. The quotient $G/H$ must be isomorphic to $mathbb{Z}_m$ for some odd $m$, since the image of $a$ under the map $G rightarrow G/H$ maps to a generator of $G/H$ and $a$ has odd order. The conjugation action of $a$ on $H$ hence gives rise to a homomorphism
$$ mathbb{Z}_m rightarrow mathrm{Aut}(H): 1 mapsto varphi=( b mapsto aba^{-1}=b^{-1}) $$
Note that $varphi$ is either trivial or of order two. Since $mvarphi =0$ for $m$ odd, it follows that $varphi$ is trivial. Hence $b=b^{-1}$, i.e. $b^2=e$.
edited Jan 14 at 19:21
Shaun
9,221113684
answered Jan 14 at 19:20
MortenMorten
605
$endgroup$
add a comment |
$begingroup$
Your proof is fine. Here is an alternative one using a bit more group theory.
Let $G$ denote the subgroup generated by $a$ and $b$. Let $H cong mathbb{Z}_n$ be the subgroup of $G$ generated by $b$. Since $aba^{-1}=b^{-1}$, it follows that $H$ is normal $G$. The quotient $G/H$ must be isomorphic to $mathbb{Z}_m$ for some odd $m$, since the image of $a$ under the map $G rightarrow G/H$ maps to a generator of $G/H$ and $a$ has odd order. The conjugation action of $a$ on $H$ hence gives rise to a homomorphism
$$ mathbb{Z}_m rightarrow mathrm{Aut}(H): 1 mapsto varphi=( b mapsto aba^{-1}=b^{-1}) $$
Note that $varphi$ is either trivial or of order two. Since $mvarphi =0$ for $m$ odd, it follows that $varphi$ is trivial. Hence $b=b^{-1}$, i.e. $b^2=e$.
edited Jan 14 at 19:21
Shaun
9,221113684
answered Jan 14 at 19:20
MortenMorten
605
$endgroup$
add a comment |
$begingroup$
Your proof is fine. Here is an alternative one using a bit more group theory.
Let $G$ denote the subgroup generated by $a$ and $b$. Let $H cong mathbb{Z}_n$ be the subgroup of $G$ generated by $b$. Since $aba^{-1}=b^{-1}$, it follows that $H$ is normal $G$. The quotient $G/H$ must be isomorphic to $mathbb{Z}_m$ for some odd $m$, since the image of $a$ under the map $G rightarrow G/H$ maps to a generator of $G/H$ and $a$ has odd order. The conjugation action of $a$ on $H$ hence gives rise to a homomorphism
$$ mathbb{Z}_m rightarrow mathrm{Aut}(H): 1 mapsto varphi=( b mapsto aba^{-1}=b^{-1}) $$
Note that $varphi$ is either trivial or of order two. Since $mvarphi =0$ for $m$ odd, it follows that $varphi$ is trivial. Hence $b=b^{-1}$, i.e. $b^2=e$.
edited Jan 14 at 19:21
Shaun
9,221113684
answered Jan 14 at 19:20
MortenMorten
605
$endgroup$
Your proof is fine. Here is an alternative one using a bit more group theory.
Let $G$ denote the subgroup generated by $a$ and $b$. Let $H cong mathbb{Z}_n$ be the subgroup of $G$ generated by $b$. Since $aba^{-1}=b^{-1}$, it follows that $H$ is normal $G$. The quotient $G/H$ must be isomorphic to $mathbb{Z}_m$ for some odd $m$, since the image of $a$ under the map $G rightarrow G/H$ maps to a generator of $G/H$ and $a$ has odd order. The conjugation action of $a$ on $H$ hence gives rise to a homomorphism
$$ mathbb{Z}_m rightarrow mathrm{Aut}(H): 1 mapsto varphi=( b mapsto aba^{-1}=b^{-1}) $$
Note that $varphi$ is either trivial or of order two. Since $mvarphi =0$ for $m$ odd, it follows that $varphi$ is trivial. Hence $b=b^{-1}$, i.e. $b^2=e$.
edited Jan 14 at 19:21
Shaun
9,221113684
answered Jan 14 at 19:20
MortenMorten
605
edited Jan 14 at 19:21
Shaun
9,221113684
edited Jan 14 at 19:21
Shaun
9,221113684
edited Jan 14 at 19:21
Shaun
9,221113684
9,221113684
answered Jan 14 at 19:20
MortenMorten
605
answered Jan 14 at 19:20
MortenMorten
605
answered Jan 14 at 19:20
MortenMorten
605
605
add a comment |
add a comment |
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3
$begingroup$
Yes the proof is OK.
$endgroup$
– Derek Holt
Jan 14 at 16:37
2
$begingroup$
Yes it is perfect
$endgroup$
– greedoid
Jan 14 at 16:38