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Elements of odd order generate a proper subgroup of a group

Proof of Lagrange theorem - Order of a subgroup divides order of the groupis there group such that has finite maximal subgroup and index of all proper subgroup of $G$ in $G$ be infinite?Can icosahedral group have subgroup of order 30?Show that if a group $G$ has odd order, any subgroup $H$ of index 3 in $G$ is normal in $G$?Quotient group with normal subgroup dividing the order of another grouphow are group order and element order related?Order of normal subgroupA group of order $p^nm$ isn't simpleIs G/H a group even if H is a subgroup of G but not a normal subgroup?Subgroup generated by all elements with odd order

3

$begingroup$

I am stuck with the following question:

Given is a group $G$ with a subgroup $H$ of index $2$, so $left [ G:H right ]=2$. I have to show that the elements of odd order of $G$ generate a proper subgroup.

What i know is that $H$ as a subgroup of index $2$ has only $2$ left cosets (and also right too). So i know that this subgroup is normal in $G$. What can i do with the order? Can anybody help me with this exercise, please?
Thank you in advance!

abstract-algebra group-theory

share|cite|improve this question

edited Feb 23 at 2:14

B.Swan

1,1001721

asked Sep 28 '15 at 21:15

LullabyLullaby

907718

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Is $G$ abelian? For a general group the odd order elements do not form a subgroup!
  $endgroup$
  – Espen Nielsen
  Sep 28 '15 at 21:22
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean that the elements of odd order generate a proper subgroup?
  $endgroup$
  – Mark Bennet
  Sep 28 '15 at 21:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Explicitly, in $S_4$, the symmetric group on $1, 2, 3, 4$, the elements (in cycle notation) $(1,2,3)$ and $(2,3,4)$ each have order $3$, but their product (composing from left to right) is $(1,3)(2,4)$, which has order $2$. The group $S_4$ does have an index $2$ subgroup, namely $A_4$, but $S_4$ is not abelian.
  $endgroup$
  – Sammy Black
  Sep 28 '15 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkBennet That's true, but since we may always add a factor $mathbbZ/2mathbbZ$, this is not a difficult requirement to meet.
  $endgroup$
  – Espen Nielsen
  Sep 28 '15 at 21:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  As has been pointed out, the result is not true, so you there must either be a mistake in the question, or you have not copied it correctly. It is true that the elements of odd order generate a proper subgroup, because all usch elements are contained in $H$.
  $endgroup$
  – Derek Holt
  Sep 29 '15 at 7:40
  

  
  
  
  

 | 
show 3 more comments

3

$begingroup$

I am stuck with the following question:

Given is a group $G$ with a subgroup $H$ of index $2$, so $left [ G:H right ]=2$. I have to show that the elements of odd order of $G$ generate a proper subgroup.

What i know is that $H$ as a subgroup of index $2$ has only $2$ left cosets (and also right too). So i know that this subgroup is normal in $G$. What can i do with the order? Can anybody help me with this exercise, please?
Thank you in advance!

abstract-algebra group-theory

share|cite|improve this question

edited Feb 23 at 2:14

B.Swan

1,1001721

asked Sep 28 '15 at 21:15

LullabyLullaby

907718

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Is $G$ abelian? For a general group the odd order elements do not form a subgroup!
  $endgroup$
  – Espen Nielsen
  Sep 28 '15 at 21:22
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean that the elements of odd order generate a proper subgroup?
  $endgroup$
  – Mark Bennet
  Sep 28 '15 at 21:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Explicitly, in $S_4$, the symmetric group on $1, 2, 3, 4$, the elements (in cycle notation) $(1,2,3)$ and $(2,3,4)$ each have order $3$, but their product (composing from left to right) is $(1,3)(2,4)$, which has order $2$. The group $S_4$ does have an index $2$ subgroup, namely $A_4$, but $S_4$ is not abelian.
  $endgroup$
  – Sammy Black
  Sep 28 '15 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkBennet That's true, but since we may always add a factor $mathbbZ/2mathbbZ$, this is not a difficult requirement to meet.
  $endgroup$
  – Espen Nielsen
  Sep 28 '15 at 21:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  As has been pointed out, the result is not true, so you there must either be a mistake in the question, or you have not copied it correctly. It is true that the elements of odd order generate a proper subgroup, because all usch elements are contained in $H$.
  $endgroup$
  – Derek Holt
  Sep 29 '15 at 7:40
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

I am stuck with the following question:

Given is a group $G$ with a subgroup $H$ of index $2$, so $left [ G:H right ]=2$. I have to show that the elements of odd order of $G$ generate a proper subgroup.

What i know is that $H$ as a subgroup of index $2$ has only $2$ left cosets (and also right too). So i know that this subgroup is normal in $G$. What can i do with the order? Can anybody help me with this exercise, please?
Thank you in advance!

abstract-algebra group-theory

share|cite|improve this question

edited Feb 23 at 2:14

B.Swan

1,1001721

asked Sep 28 '15 at 21:15

LullabyLullaby

907718

$endgroup$

share|cite|improve this question

edited Feb 23 at 2:14

B.Swan

1,1001721

asked Sep 28 '15 at 21:15

LullabyLullaby

907718

share|cite|improve this question

edited Feb 23 at 2:14

B.Swan

1,1001721

asked Sep 28 '15 at 21:15

LullabyLullaby

907718

edited Feb 23 at 2:14

B.Swan

1,1001721

edited Feb 23 at 2:14

B.Swan

1,1001721

asked Sep 28 '15 at 21:15

LullabyLullaby

907718

asked Sep 28 '15 at 21:15

LullabyLullaby

907718

1 Answer
1

active

oldest

votes

1

$begingroup$

The elements of odd order generate a subgroup, let's call it $P$, because
$1 in G$ has order $1$, which is odd. So $1 in P$.

Thus $P$ contains some elements and since we are generating a group, we automatically generate the inverses and all products, so $P$ will be a subgroup.

---

We still need to show that $Pneq G$ for $P$ to be a proper subgroup.

We know that there is an $H leq G$ with $[G:H]=2$, so $Hneq G$.

We will show that $P$ lies in $H$.

Theorem 2.
If $G$ is a finite group and $N triangleleft G$ then any element of $G$ with order relatively prime to $[G:N]$ lies in $N$. In
particular, if $N$ has index $2$ then all elements of $ G$ with odd
order lie in $N$.

Proof: Let $g$ be an element of $G$ with order $m$, which is relatively prime to $[G:N]$. The equation $g^m=e$ gives
$(gN)^m=N in G/N$. Also $(gN)^[G:N]=N$, as $[G:N]$ is the order
of $G/N$.

So the order of $gN in G/N$ divides $m$ and $[G:N]$.

These numbers are relatively prime, so $gN=N$, which means $g in N$.

Now we know that all odd elements lie in $H$. Since $H$ is a subgroup, it is closed, so the whole subgroup generated by odd elements must lie in $H$. Thus $Pneq G$.

share|cite|improve this answer

edited 2 days ago

answered Feb 23 at 0:09

B.SwanB.Swan

1,1001721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Most of this could probably be left out since the statement that the elements generate a subgroup is vacuously true since this holds for any subset. On the other hand, including a short proof of the main result cited would make the answer self-contained which is generally preferable.
  $endgroup$
  – Tobias Kildetoft
  Feb 23 at 7:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, I will edit my answer accordingly.
  $endgroup$
  – B.Swan
  Feb 23 at 16:53
  

  
  
  
  

add a comment | 

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1

$begingroup$

The elements of odd order generate a subgroup, let's call it $P$, because
$1 in G$ has order $1$, which is odd. So $1 in P$.

Thus $P$ contains some elements and since we are generating a group, we automatically generate the inverses and all products, so $P$ will be a subgroup.

---

We still need to show that $Pneq G$ for $P$ to be a proper subgroup.

We know that there is an $H leq G$ with $[G:H]=2$, so $Hneq G$.

We will show that $P$ lies in $H$.

Theorem 2.
If $G$ is a finite group and $N triangleleft G$ then any element of $G$ with order relatively prime to $[G:N]$ lies in $N$. In
particular, if $N$ has index $2$ then all elements of $ G$ with odd
order lie in $N$.

Proof: Let $g$ be an element of $G$ with order $m$, which is relatively prime to $[G:N]$. The equation $g^m=e$ gives
$(gN)^m=N in G/N$. Also $(gN)^[G:N]=N$, as $[G:N]$ is the order
of $G/N$.

So the order of $gN in G/N$ divides $m$ and $[G:N]$.

These numbers are relatively prime, so $gN=N$, which means $g in N$.

Now we know that all odd elements lie in $H$. Since $H$ is a subgroup, it is closed, so the whole subgroup generated by odd elements must lie in $H$. Thus $Pneq G$.

share|cite|improve this answer

edited 2 days ago

answered Feb 23 at 0:09

B.SwanB.Swan

1,1001721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Most of this could probably be left out since the statement that the elements generate a subgroup is vacuously true since this holds for any subset. On the other hand, including a short proof of the main result cited would make the answer self-contained which is generally preferable.
  $endgroup$
  – Tobias Kildetoft
  Feb 23 at 7:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, I will edit my answer accordingly.
  $endgroup$
  – B.Swan
  Feb 23 at 16:53
  

  
  
  
  

add a comment | 

1

$begingroup$

The elements of odd order generate a subgroup, let's call it $P$, because
$1 in G$ has order $1$, which is odd. So $1 in P$.

Thus $P$ contains some elements and since we are generating a group, we automatically generate the inverses and all products, so $P$ will be a subgroup.

---

We still need to show that $Pneq G$ for $P$ to be a proper subgroup.

We know that there is an $H leq G$ with $[G:H]=2$, so $Hneq G$.

We will show that $P$ lies in $H$.

Theorem 2.
If $G$ is a finite group and $N triangleleft G$ then any element of $G$ with order relatively prime to $[G:N]$ lies in $N$. In
particular, if $N$ has index $2$ then all elements of $ G$ with odd
order lie in $N$.

Proof: Let $g$ be an element of $G$ with order $m$, which is relatively prime to $[G:N]$. The equation $g^m=e$ gives
$(gN)^m=N in G/N$. Also $(gN)^[G:N]=N$, as $[G:N]$ is the order
of $G/N$.

So the order of $gN in G/N$ divides $m$ and $[G:N]$.

These numbers are relatively prime, so $gN=N$, which means $g in N$.

Now we know that all odd elements lie in $H$. Since $H$ is a subgroup, it is closed, so the whole subgroup generated by odd elements must lie in $H$. Thus $Pneq G$.

share|cite|improve this answer

edited 2 days ago

answered Feb 23 at 0:09

B.SwanB.Swan

1,1001721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Most of this could probably be left out since the statement that the elements generate a subgroup is vacuously true since this holds for any subset. On the other hand, including a short proof of the main result cited would make the answer self-contained which is generally preferable.
  $endgroup$
  – Tobias Kildetoft
  Feb 23 at 7:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, I will edit my answer accordingly.
  $endgroup$
  – B.Swan
  Feb 23 at 16:53
  

  
  
  
  

add a comment | 

$begingroup$

The elements of odd order generate a subgroup, let's call it $P$, because
$1 in G$ has order $1$, which is odd. So $1 in P$.

Thus $P$ contains some elements and since we are generating a group, we automatically generate the inverses and all products, so $P$ will be a subgroup.

---

We still need to show that $Pneq G$ for $P$ to be a proper subgroup.

We know that there is an $H leq G$ with $[G:H]=2$, so $Hneq G$.

We will show that $P$ lies in $H$.

Theorem 2.
If $G$ is a finite group and $N triangleleft G$ then any element of $G$ with order relatively prime to $[G:N]$ lies in $N$. In
particular, if $N$ has index $2$ then all elements of $ G$ with odd
order lie in $N$.

Proof: Let $g$ be an element of $G$ with order $m$, which is relatively prime to $[G:N]$. The equation $g^m=e$ gives
$(gN)^m=N in G/N$. Also $(gN)^[G:N]=N$, as $[G:N]$ is the order
of $G/N$.

So the order of $gN in G/N$ divides $m$ and $[G:N]$.

These numbers are relatively prime, so $gN=N$, which means $g in N$.

Now we know that all odd elements lie in $H$. Since $H$ is a subgroup, it is closed, so the whole subgroup generated by odd elements must lie in $H$. Thus $Pneq G$.

share|cite|improve this answer

edited 2 days ago

answered Feb 23 at 0:09

B.SwanB.Swan

1,1001721

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered Feb 23 at 0:09

B.SwanB.Swan

1,1001721

answered Feb 23 at 0:09

B.SwanB.Swan

1,1001721

answered Feb 23 at 0:09

B.SwanB.Swan

1,1001721