Mixing
An abelian group $G$ of order $35$ with $g^{35}=e$ for all $gin G$ is cyclic.
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.20.
Suppose that $G$ is an Abelian group of order $35$ and every element of $G$ satisfies the equation $x^{35}=e$. Prove that $G$ is cyclic. Does your argument work if $35$ is replaced with $33$?
There's definitely something I don't understand here. I am/was under the impression that, for any group $H$ and any $hin H$, we have $h^{lvert Hrvert}=e$; indeed: the cyclic subgroup $langle hrangle$ of $H$ has the same order as a group as the order $lvert hrvert$ of $h$ as an element of $H$; Lagrange's theorem then gives that $lvert hrvert$ divides $lvert Hrvert$, so that then $h^{lvert Hrvert}=e$.
So what gives?
I get the feeling that it's something obvious.
group-theory finite-groups abelian-groups cyclic-groups
asked Jan 13 at 19:10
ShaunShaun
9,182113684
$endgroup$
add a comment |
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.20.
Suppose that $G$ is an Abelian group of order $35$ and every element of $G$ satisfies the equation $x^{35}=e$. Prove that $G$ is cyclic. Does your argument work if $35$ is replaced with $33$?
There's definitely something I don't understand here. I am/was under the impression that, for any group $H$ and any $hin H$, we have $h^{lvert Hrvert}=e$; indeed: the cyclic subgroup $langle hrangle$ of $H$ has the same order as a group as the order $lvert hrvert$ of $h$ as an element of $H$; Lagrange's theorem then gives that $lvert hrvert$ divides $lvert Hrvert$, so that then $h^{lvert Hrvert}=e$.
So what gives?
I get the feeling that it's something obvious.
group-theory finite-groups abelian-groups cyclic-groups
asked Jan 13 at 19:10
ShaunShaun
9,182113684
$endgroup$
3
$begingroup$
It is a very odd question. It's true because every group of order $35$ is cyclic, but the condition applies to a group of order $35$ whether we know that or not, like you said.
$endgroup$
– Matt Samuel
Jan 13 at 19:14
1
$begingroup$
It is already enough that the group order is $35$ (or $33$). We only have the cyclic (and therefore abelian) group in this case. If a number is squarefree and has two prime factors $p$ and $q$ ($p<q$) , there is only one group with this order if and only if $p$ does not divide $q-1$
$endgroup$
– Peter
Jan 13 at 19:15
add a comment |
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.20.
Suppose that $G$ is an Abelian group of order $35$ and every element of $G$ satisfies the equation $x^{35}=e$. Prove that $G$ is cyclic. Does your argument work if $35$ is replaced with $33$?
There's definitely something I don't understand here. I am/was under the impression that, for any group $H$ and any $hin H$, we have $h^{lvert Hrvert}=e$; indeed: the cyclic subgroup $langle hrangle$ of $H$ has the same order as a group as the order $lvert hrvert$ of $h$ as an element of $H$; Lagrange's theorem then gives that $lvert hrvert$ divides $lvert Hrvert$, so that then $h^{lvert Hrvert}=e$.
So what gives?
I get the feeling that it's something obvious.
group-theory finite-groups abelian-groups cyclic-groups
asked Jan 13 at 19:10
ShaunShaun
9,182113684
$endgroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.20.
Suppose that $G$ is an Abelian group of order $35$ and every element of $G$ satisfies the equation $x^{35}=e$. Prove that $G$ is cyclic. Does your argument work if $35$ is replaced with $33$?
There's definitely something I don't understand here. I am/was under the impression that, for any group $H$ and any $hin H$, we have $h^{lvert Hrvert}=e$; indeed: the cyclic subgroup $langle hrangle$ of $H$ has the same order as a group as the order $lvert hrvert$ of $h$ as an element of $H$; Lagrange's theorem then gives that $lvert hrvert$ divides $lvert Hrvert$, so that then $h^{lvert Hrvert}=e$.
So what gives?
I get the feeling that it's something obvious.
group-theory finite-groups abelian-groups cyclic-groups
group-theory finite-groups abelian-groups cyclic-groups
asked Jan 13 at 19:10
ShaunShaun
9,182113684
asked Jan 13 at 19:10
ShaunShaun
9,182113684
edited Jan 13 at 21:58
Shaun
asked Jan 13 at 19:10
ShaunShaun
9,182113684
asked Jan 13 at 19:10
ShaunShaun
9,182113684
asked Jan 13 at 19:10
ShaunShaun
9,182113684
9,182113684
3
$begingroup$
It is a very odd question. It's true because every group of order $35$ is cyclic, but the condition applies to a group of order $35$ whether we know that or not, like you said.
$endgroup$
– Matt Samuel
Jan 13 at 19:14
1
$begingroup$
It is already enough that the group order is $35$ (or $33$). We only have the cyclic (and therefore abelian) group in this case. If a number is squarefree and has two prime factors $p$ and $q$ ($p<q$) , there is only one group with this order if and only if $p$ does not divide $q-1$
$endgroup$
– Peter
Jan 13 at 19:15
add a comment |
3
$begingroup$
It is a very odd question. It's true because every group of order $35$ is cyclic, but the condition applies to a group of order $35$ whether we know that or not, like you said.
$endgroup$
– Matt Samuel
Jan 13 at 19:14
1
$begingroup$
It is already enough that the group order is $35$ (or $33$). We only have the cyclic (and therefore abelian) group in this case. If a number is squarefree and has two prime factors $p$ and $q$ ($p<q$) , there is only one group with this order if and only if $p$ does not divide $q-1$
$endgroup$
– Peter
Jan 13 at 19:15
3
3
$begingroup$
It is a very odd question. It's true because every group of order $35$ is cyclic, but the condition applies to a group of order $35$ whether we know that or not, like you said.
$endgroup$
– Matt Samuel
Jan 13 at 19:14
$begingroup$
It is a very odd question. It's true because every group of order $35$ is cyclic, but the condition applies to a group of order $35$ whether we know that or not, like you said.
$endgroup$
– Matt Samuel
Jan 13 at 19:14
1
1
$begingroup$
It is already enough that the group order is $35$ (or $33$). We only have the cyclic (and therefore abelian) group in this case. If a number is squarefree and has two prime factors $p$ and $q$ ($p<q$) , there is only one group with this order if and only if $p$ does not divide $q-1$
$endgroup$
– Peter
Jan 13 at 19:15
$begingroup$
It is already enough that the group order is $35$ (or $33$). We only have the cyclic (and therefore abelian) group in this case. If a number is squarefree and has two prime factors $p$ and $q$ ($p<q$) , there is only one group with this order if and only if $p$ does not divide $q-1$
$endgroup$
– Peter
Jan 13 at 19:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
My guess is that Gallian only proves Lagrange's theorem later.
If $(forall gin G):g^{35}=e$, then the order of every element of $G$ is $1$, $5$, $7$, or $35$. Of course, the goal is to prove that some element of $G$ has order $35$.
Suppose that every element other than $e$ has order $5$. Then every element of $Gsetminus{e}$ belongs to sone subgroup of order $5$. Each such subgroup will consist of $4$ elements of order $5$ plus $e$. But there's a problem here: $35$ is not of the form $4k+1$. So, some element from $Gsetminus{e}$ must have order $7$ or $35$. And, by the same argument, not all elements from $Gsetminus{e}$ have order $7$.
Therefore, some $ain G$ has order $5$ or $35$ and some $bin Gsetminus{e}$ has order $7$ or $35$. If one of them has order $35$, we're done: $G$ is cyclic. Otherwise, $ab$ has order $35$ and, again, $G$ is cyclic.
And, clearly, this argument does not apply to $33$.
answered Jan 13 at 19:28
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
You're right: Gallian hasn't introduced Lagrange's Theorem yet. Your proof is lovely. Thank you :)
$endgroup$
– Shaun
Jan 13 at 19:34
add a comment |
$begingroup$
If $G$ is a group of order $n$, then $x^n=e$ by Euler's theorem for all $xin G$. So this is an empty condition which is satisfied anyway. The point is that groups of order $pq$ are classified for primes $p<q$, depending on whether or not $pmid q-1$. In one case all such groups are cyclic, in the other case there is a non-abelian group of that order.
Now check $(p,q)=(5,7),(3,11)$.
Reference: Let $p<q$ be distinct prime numbers and $G$ be a group with $|G|=pq$
answered Jan 13 at 19:13
Dietrich BurdeDietrich Burde
79.1k647103
$endgroup$
add a comment |
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$begingroup$
My guess is that Gallian only proves Lagrange's theorem later.
If $(forall gin G):g^{35}=e$, then the order of every element of $G$ is $1$, $5$, $7$, or $35$. Of course, the goal is to prove that some element of $G$ has order $35$.
Suppose that every element other than $e$ has order $5$. Then every element of $Gsetminus{e}$ belongs to sone subgroup of order $5$. Each such subgroup will consist of $4$ elements of order $5$ plus $e$. But there's a problem here: $35$ is not of the form $4k+1$. So, some element from $Gsetminus{e}$ must have order $7$ or $35$. And, by the same argument, not all elements from $Gsetminus{e}$ have order $7$.
Therefore, some $ain G$ has order $5$ or $35$ and some $bin Gsetminus{e}$ has order $7$ or $35$. If one of them has order $35$, we're done: $G$ is cyclic. Otherwise, $ab$ has order $35$ and, again, $G$ is cyclic.
And, clearly, this argument does not apply to $33$.
answered Jan 13 at 19:28
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
You're right: Gallian hasn't introduced Lagrange's Theorem yet. Your proof is lovely. Thank you :)
$endgroup$
– Shaun
Jan 13 at 19:34
add a comment |
$begingroup$
My guess is that Gallian only proves Lagrange's theorem later.
If $(forall gin G):g^{35}=e$, then the order of every element of $G$ is $1$, $5$, $7$, or $35$. Of course, the goal is to prove that some element of $G$ has order $35$.
Suppose that every element other than $e$ has order $5$. Then every element of $Gsetminus{e}$ belongs to sone subgroup of order $5$. Each such subgroup will consist of $4$ elements of order $5$ plus $e$. But there's a problem here: $35$ is not of the form $4k+1$. So, some element from $Gsetminus{e}$ must have order $7$ or $35$. And, by the same argument, not all elements from $Gsetminus{e}$ have order $7$.
Therefore, some $ain G$ has order $5$ or $35$ and some $bin Gsetminus{e}$ has order $7$ or $35$. If one of them has order $35$, we're done: $G$ is cyclic. Otherwise, $ab$ has order $35$ and, again, $G$ is cyclic.
And, clearly, this argument does not apply to $33$.
answered Jan 13 at 19:28
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
You're right: Gallian hasn't introduced Lagrange's Theorem yet. Your proof is lovely. Thank you :)
$endgroup$
– Shaun
Jan 13 at 19:34
add a comment |
$begingroup$
My guess is that Gallian only proves Lagrange's theorem later.
If $(forall gin G):g^{35}=e$, then the order of every element of $G$ is $1$, $5$, $7$, or $35$. Of course, the goal is to prove that some element of $G$ has order $35$.
Suppose that every element other than $e$ has order $5$. Then every element of $Gsetminus{e}$ belongs to sone subgroup of order $5$. Each such subgroup will consist of $4$ elements of order $5$ plus $e$. But there's a problem here: $35$ is not of the form $4k+1$. So, some element from $Gsetminus{e}$ must have order $7$ or $35$. And, by the same argument, not all elements from $Gsetminus{e}$ have order $7$.
Therefore, some $ain G$ has order $5$ or $35$ and some $bin Gsetminus{e}$ has order $7$ or $35$. If one of them has order $35$, we're done: $G$ is cyclic. Otherwise, $ab$ has order $35$ and, again, $G$ is cyclic.
And, clearly, this argument does not apply to $33$.
answered Jan 13 at 19:28
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
My guess is that Gallian only proves Lagrange's theorem later.
If $(forall gin G):g^{35}=e$, then the order of every element of $G$ is $1$, $5$, $7$, or $35$. Of course, the goal is to prove that some element of $G$ has order $35$.
Suppose that every element other than $e$ has order $5$. Then every element of $Gsetminus{e}$ belongs to sone subgroup of order $5$. Each such subgroup will consist of $4$ elements of order $5$ plus $e$. But there's a problem here: $35$ is not of the form $4k+1$. So, some element from $Gsetminus{e}$ must have order $7$ or $35$. And, by the same argument, not all elements from $Gsetminus{e}$ have order $7$.
Therefore, some $ain G$ has order $5$ or $35$ and some $bin Gsetminus{e}$ has order $7$ or $35$. If one of them has order $35$, we're done: $G$ is cyclic. Otherwise, $ab$ has order $35$ and, again, $G$ is cyclic.
And, clearly, this argument does not apply to $33$.
answered Jan 13 at 19:28
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 19:28
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 19:28
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 19:28
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
$begingroup$
You're right: Gallian hasn't introduced Lagrange's Theorem yet. Your proof is lovely. Thank you :)
$endgroup$
– Shaun
Jan 13 at 19:34
add a comment |
$begingroup$
You're right: Gallian hasn't introduced Lagrange's Theorem yet. Your proof is lovely. Thank you :)
$endgroup$
– Shaun
Jan 13 at 19:34
$begingroup$
You're right: Gallian hasn't introduced Lagrange's Theorem yet. Your proof is lovely. Thank you :)
$endgroup$
– Shaun
Jan 13 at 19:34
$begingroup$
You're right: Gallian hasn't introduced Lagrange's Theorem yet. Your proof is lovely. Thank you :)
$endgroup$
– Shaun
Jan 13 at 19:34
add a comment |
$begingroup$
If $G$ is a group of order $n$, then $x^n=e$ by Euler's theorem for all $xin G$. So this is an empty condition which is satisfied anyway. The point is that groups of order $pq$ are classified for primes $p<q$, depending on whether or not $pmid q-1$. In one case all such groups are cyclic, in the other case there is a non-abelian group of that order.
Now check $(p,q)=(5,7),(3,11)$.
Reference: Let $p<q$ be distinct prime numbers and $G$ be a group with $|G|=pq$
answered Jan 13 at 19:13
Dietrich BurdeDietrich Burde
79.1k647103
$endgroup$
add a comment |
$begingroup$
If $G$ is a group of order $n$, then $x^n=e$ by Euler's theorem for all $xin G$. So this is an empty condition which is satisfied anyway. The point is that groups of order $pq$ are classified for primes $p<q$, depending on whether or not $pmid q-1$. In one case all such groups are cyclic, in the other case there is a non-abelian group of that order.
Now check $(p,q)=(5,7),(3,11)$.
Reference: Let $p<q$ be distinct prime numbers and $G$ be a group with $|G|=pq$
answered Jan 13 at 19:13
Dietrich BurdeDietrich Burde
79.1k647103
$endgroup$
add a comment |
$begingroup$
If $G$ is a group of order $n$, then $x^n=e$ by Euler's theorem for all $xin G$. So this is an empty condition which is satisfied anyway. The point is that groups of order $pq$ are classified for primes $p<q$, depending on whether or not $pmid q-1$. In one case all such groups are cyclic, in the other case there is a non-abelian group of that order.
Now check $(p,q)=(5,7),(3,11)$.
Reference: Let $p<q$ be distinct prime numbers and $G$ be a group with $|G|=pq$
answered Jan 13 at 19:13
Dietrich BurdeDietrich Burde
79.1k647103
$endgroup$
If $G$ is a group of order $n$, then $x^n=e$ by Euler's theorem for all $xin G$. So this is an empty condition which is satisfied anyway. The point is that groups of order $pq$ are classified for primes $p<q$, depending on whether or not $pmid q-1$. In one case all such groups are cyclic, in the other case there is a non-abelian group of that order.
Now check $(p,q)=(5,7),(3,11)$.
Reference: Let $p<q$ be distinct prime numbers and $G$ be a group with $|G|=pq$
answered Jan 13 at 19:13
Dietrich BurdeDietrich Burde
79.1k647103
answered Jan 13 at 19:13
Dietrich BurdeDietrich Burde
79.1k647103
answered Jan 13 at 19:13
Dietrich BurdeDietrich Burde
79.1k647103
answered Jan 13 at 19:13
Dietrich BurdeDietrich Burde
79.1k647103
79.1k647103
add a comment |
add a comment |
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$begingroup$
It is a very odd question. It's true because every group of order $35$ is cyclic, but the condition applies to a group of order $35$ whether we know that or not, like you said.
$endgroup$
– Matt Samuel
Jan 13 at 19:14
1
$begingroup$
It is already enough that the group order is $35$ (or $33$). We only have the cyclic (and therefore abelian) group in this case. If a number is squarefree and has two prime factors $p$ and $q$ ($p<q$) , there is only one group with this order if and only if $p$ does not divide $q-1$
$endgroup$
– Peter
Jan 13 at 19:15