Mixing
Let $G$ be a finite group of even order. Then which of the following statements is correct?
$begingroup$
Let $G$ be a finite group of even order. Then which of the following
statements is correct ?
$1.$ The number of elements of order $2$ in $G$ is even
$2.$ The number of elements of order $2$ in $G$ is odd
$3.$ G has no subgroup of order $2$
$4.$ None of the above.
My attempt : I take $G= K_4 $, then option $1$ is correct
Is it true ?
Any hints/solution
Thank you!
abstract-algebra
edited Jan 5 at 15:57
Chinnapparaj R
5,3341828
asked Jan 5 at 15:29
jasminejasmine
1,689416
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group of even order. Then which of the following
statements is correct ?
$1.$ The number of elements of order $2$ in $G$ is even
$2.$ The number of elements of order $2$ in $G$ is odd
$3.$ G has no subgroup of order $2$
$4.$ None of the above.
My attempt : I take $G= K_4 $, then option $1$ is correct
Is it true ?
Any hints/solution
Thank you!
abstract-algebra
edited Jan 5 at 15:57
Chinnapparaj R
5,3341828
asked Jan 5 at 15:29
jasminejasmine
1,689416
$endgroup$
2
$begingroup$
Whatever $K_4$ is, it isn't every possible group of even order.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 15:33
1
$begingroup$
I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
$endgroup$
– paw88789
Jan 5 at 15:34
$begingroup$
@LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
$endgroup$
– jasmine
Jan 5 at 15:34
add a comment |
$begingroup$
Let $G$ be a finite group of even order. Then which of the following
statements is correct ?
$1.$ The number of elements of order $2$ in $G$ is even
$2.$ The number of elements of order $2$ in $G$ is odd
$3.$ G has no subgroup of order $2$
$4.$ None of the above.
My attempt : I take $G= K_4 $, then option $1$ is correct
Is it true ?
Any hints/solution
Thank you!
abstract-algebra
edited Jan 5 at 15:57
Chinnapparaj R
5,3341828
asked Jan 5 at 15:29
jasminejasmine
1,689416
$endgroup$
Let $G$ be a finite group of even order. Then which of the following
statements is correct ?
$1.$ The number of elements of order $2$ in $G$ is even
$2.$ The number of elements of order $2$ in $G$ is odd
$3.$ G has no subgroup of order $2$
$4.$ None of the above.
My attempt : I take $G= K_4 $, then option $1$ is correct
Is it true ?
Any hints/solution
Thank you!
abstract-algebra
abstract-algebra
edited Jan 5 at 15:57
Chinnapparaj R
5,3341828
asked Jan 5 at 15:29
jasminejasmine
1,689416
edited Jan 5 at 15:57
Chinnapparaj R
5,3341828
asked Jan 5 at 15:29
jasminejasmine
1,689416
edited Jan 5 at 15:57
Chinnapparaj R
5,3341828
edited Jan 5 at 15:57
Chinnapparaj R
5,3341828
edited Jan 5 at 15:57
Chinnapparaj R
5,3341828
5,3341828
asked Jan 5 at 15:29
jasminejasmine
1,689416
asked Jan 5 at 15:29
jasminejasmine
1,689416
asked Jan 5 at 15:29
jasminejasmine
1,689416
1,689416
2
$begingroup$
Whatever $K_4$ is, it isn't every possible group of even order.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 15:33
1
$begingroup$
I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
$endgroup$
– paw88789
Jan 5 at 15:34
$begingroup$
@LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
$endgroup$
– jasmine
Jan 5 at 15:34
add a comment |
2
$begingroup$
Whatever $K_4$ is, it isn't every possible group of even order.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 15:33
1
$begingroup$
I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
$endgroup$
– paw88789
Jan 5 at 15:34
$begingroup$
@LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
$endgroup$
– jasmine
Jan 5 at 15:34
2
2
$begingroup$
Whatever $K_4$ is, it isn't every possible group of even order.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 15:33
$begingroup$
Whatever $K_4$ is, it isn't every possible group of even order.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 15:33
1
1
$begingroup$
I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
$endgroup$
– paw88789
Jan 5 at 15:34
$begingroup$
I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
$endgroup$
– paw88789
Jan 5 at 15:34
$begingroup$
@LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
$endgroup$
– jasmine
Jan 5 at 15:34
$begingroup$
@LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
$endgroup$
– jasmine
Jan 5 at 15:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Option 1 is false, actually the group $K_4$ has $3$ elements of order $2$
The group $G$ must has an element of order $2$ is true, and it is an easy exercise to you!
If $a neq a^{-1}$, then $vert a vert$ is not $2$, so this type of elements comes in pairs, which make an even count. Call this count as $k$, then
$$vert G vert =underbrace{vert {e} vert}_{1} + underbrace{vert {a in G :a neq a^{-1} } vert}_{k;; text{an even count}} +text{number of remaining order $2$ elements}$$
But Whole count is even, so the last count must be ODD!
Option $3$ is obviously false!
answered Jan 5 at 15:35
Chinnapparaj RChinnapparaj R
5,3341828
$endgroup$
$begingroup$
im not getting why u write $ |G|= 1 + k $?
$endgroup$
– jasmine
Jan 5 at 15:40
1
$begingroup$
That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
$endgroup$
– Chinnapparaj R
Jan 5 at 15:43
add a comment |
$begingroup$
As given that group $G$ is of even order so number of elements of order $2$ should be odd since any element of order $2$ is self inverse and any other element of order more than $2$ must appear in the group with its inverse element also so total number of elements in finite group of even order is number of self inverse elements+number of nonself inverse elements clearly number of non self order elements are even and in any group identity element is always self inverse so number of elements of order $2$ should be odd(since self inverse elements other than identity are of order $2$ only). Therefore option $1$ is false, for example you can consider $S_3$ group of symmetry on $3$ symbols it has $3$ elements of order $2$. So option $2$ s correct. Option $3$ can be easily discarded using the Syllow theorem.
answered Jan 5 at 16:13
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
497
$endgroup$
$begingroup$
Suppose $G$ is a group of even order and it does not have any element of order 2. Let $a in G^*$ then o(a)>2 so $a^{-1}$ will be distinct from $a$ and will belong to G. So it can be seen that any element of $G$ other than identity must occur in pair in $G$. So order of $G$ become odd including identity. Which leads to a contradiction. So $G$ will have an element of order 2 so a subgroup of order 2 also.
$endgroup$
– MANI SHANKAR PANDEY
Jan 5 at 17:51
add a comment |
Your Answer
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2 Answers
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2 Answers
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votes
$begingroup$
Option 1 is false, actually the group $K_4$ has $3$ elements of order $2$
The group $G$ must has an element of order $2$ is true, and it is an easy exercise to you!
If $a neq a^{-1}$, then $vert a vert$ is not $2$, so this type of elements comes in pairs, which make an even count. Call this count as $k$, then
$$vert G vert =underbrace{vert {e} vert}_{1} + underbrace{vert {a in G :a neq a^{-1} } vert}_{k;; text{an even count}} +text{number of remaining order $2$ elements}$$
But Whole count is even, so the last count must be ODD!
Option $3$ is obviously false!
answered Jan 5 at 15:35
Chinnapparaj RChinnapparaj R
5,3341828
$endgroup$
$begingroup$
im not getting why u write $ |G|= 1 + k $?
$endgroup$
– jasmine
Jan 5 at 15:40
1
$begingroup$
That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
$endgroup$
– Chinnapparaj R
Jan 5 at 15:43
add a comment |
$begingroup$
Option 1 is false, actually the group $K_4$ has $3$ elements of order $2$
The group $G$ must has an element of order $2$ is true, and it is an easy exercise to you!
If $a neq a^{-1}$, then $vert a vert$ is not $2$, so this type of elements comes in pairs, which make an even count. Call this count as $k$, then
$$vert G vert =underbrace{vert {e} vert}_{1} + underbrace{vert {a in G :a neq a^{-1} } vert}_{k;; text{an even count}} +text{number of remaining order $2$ elements}$$
But Whole count is even, so the last count must be ODD!
Option $3$ is obviously false!
answered Jan 5 at 15:35
Chinnapparaj RChinnapparaj R
5,3341828
$endgroup$
$begingroup$
im not getting why u write $ |G|= 1 + k $?
$endgroup$
– jasmine
Jan 5 at 15:40
1
$begingroup$
That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
$endgroup$
– Chinnapparaj R
Jan 5 at 15:43
add a comment |
$begingroup$
Option 1 is false, actually the group $K_4$ has $3$ elements of order $2$
The group $G$ must has an element of order $2$ is true, and it is an easy exercise to you!
If $a neq a^{-1}$, then $vert a vert$ is not $2$, so this type of elements comes in pairs, which make an even count. Call this count as $k$, then
$$vert G vert =underbrace{vert {e} vert}_{1} + underbrace{vert {a in G :a neq a^{-1} } vert}_{k;; text{an even count}} +text{number of remaining order $2$ elements}$$
But Whole count is even, so the last count must be ODD!
Option $3$ is obviously false!
answered Jan 5 at 15:35
Chinnapparaj RChinnapparaj R
5,3341828
$endgroup$
Option 1 is false, actually the group $K_4$ has $3$ elements of order $2$
The group $G$ must has an element of order $2$ is true, and it is an easy exercise to you!
If $a neq a^{-1}$, then $vert a vert$ is not $2$, so this type of elements comes in pairs, which make an even count. Call this count as $k$, then
$$vert G vert =underbrace{vert {e} vert}_{1} + underbrace{vert {a in G :a neq a^{-1} } vert}_{k;; text{an even count}} +text{number of remaining order $2$ elements}$$
But Whole count is even, so the last count must be ODD!
Option $3$ is obviously false!
answered Jan 5 at 15:35
Chinnapparaj RChinnapparaj R
5,3341828
edited Jan 5 at 15:49
answered Jan 5 at 15:35
Chinnapparaj RChinnapparaj R
5,3341828
answered Jan 5 at 15:35
Chinnapparaj RChinnapparaj R
5,3341828
answered Jan 5 at 15:35
Chinnapparaj RChinnapparaj R
5,3341828
5,3341828
$begingroup$
im not getting why u write $ |G|= 1 + k $?
$endgroup$
– jasmine
Jan 5 at 15:40
1
$begingroup$
That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
$endgroup$
– Chinnapparaj R
Jan 5 at 15:43
add a comment |
$begingroup$
im not getting why u write $ |G|= 1 + k $?
$endgroup$
– jasmine
Jan 5 at 15:40
1
$begingroup$
That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
$endgroup$
– Chinnapparaj R
Jan 5 at 15:43
$begingroup$
im not getting why u write $ |G|= 1 + k $?
$endgroup$
– jasmine
Jan 5 at 15:40
$begingroup$
im not getting why u write $ |G|= 1 + k $?
$endgroup$
– jasmine
Jan 5 at 15:40
1
1
$begingroup$
That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
$endgroup$
– Chinnapparaj R
Jan 5 at 15:43
$begingroup$
That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
$endgroup$
– Chinnapparaj R
Jan 5 at 15:43
add a comment |
$begingroup$
As given that group $G$ is of even order so number of elements of order $2$ should be odd since any element of order $2$ is self inverse and any other element of order more than $2$ must appear in the group with its inverse element also so total number of elements in finite group of even order is number of self inverse elements+number of nonself inverse elements clearly number of non self order elements are even and in any group identity element is always self inverse so number of elements of order $2$ should be odd(since self inverse elements other than identity are of order $2$ only). Therefore option $1$ is false, for example you can consider $S_3$ group of symmetry on $3$ symbols it has $3$ elements of order $2$. So option $2$ s correct. Option $3$ can be easily discarded using the Syllow theorem.
answered Jan 5 at 16:13
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
497
$endgroup$
$begingroup$
Suppose $G$ is a group of even order and it does not have any element of order 2. Let $a in G^*$ then o(a)>2 so $a^{-1}$ will be distinct from $a$ and will belong to G. So it can be seen that any element of $G$ other than identity must occur in pair in $G$. So order of $G$ become odd including identity. Which leads to a contradiction. So $G$ will have an element of order 2 so a subgroup of order 2 also.
$endgroup$
– MANI SHANKAR PANDEY
Jan 5 at 17:51
add a comment |
$begingroup$
As given that group $G$ is of even order so number of elements of order $2$ should be odd since any element of order $2$ is self inverse and any other element of order more than $2$ must appear in the group with its inverse element also so total number of elements in finite group of even order is number of self inverse elements+number of nonself inverse elements clearly number of non self order elements are even and in any group identity element is always self inverse so number of elements of order $2$ should be odd(since self inverse elements other than identity are of order $2$ only). Therefore option $1$ is false, for example you can consider $S_3$ group of symmetry on $3$ symbols it has $3$ elements of order $2$. So option $2$ s correct. Option $3$ can be easily discarded using the Syllow theorem.
answered Jan 5 at 16:13
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
497
$endgroup$
$begingroup$
Suppose $G$ is a group of even order and it does not have any element of order 2. Let $a in G^*$ then o(a)>2 so $a^{-1}$ will be distinct from $a$ and will belong to G. So it can be seen that any element of $G$ other than identity must occur in pair in $G$. So order of $G$ become odd including identity. Which leads to a contradiction. So $G$ will have an element of order 2 so a subgroup of order 2 also.
$endgroup$
– MANI SHANKAR PANDEY
Jan 5 at 17:51
add a comment |
$begingroup$
As given that group $G$ is of even order so number of elements of order $2$ should be odd since any element of order $2$ is self inverse and any other element of order more than $2$ must appear in the group with its inverse element also so total number of elements in finite group of even order is number of self inverse elements+number of nonself inverse elements clearly number of non self order elements are even and in any group identity element is always self inverse so number of elements of order $2$ should be odd(since self inverse elements other than identity are of order $2$ only). Therefore option $1$ is false, for example you can consider $S_3$ group of symmetry on $3$ symbols it has $3$ elements of order $2$. So option $2$ s correct. Option $3$ can be easily discarded using the Syllow theorem.
answered Jan 5 at 16:13
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
497
$endgroup$
As given that group $G$ is of even order so number of elements of order $2$ should be odd since any element of order $2$ is self inverse and any other element of order more than $2$ must appear in the group with its inverse element also so total number of elements in finite group of even order is number of self inverse elements+number of nonself inverse elements clearly number of non self order elements are even and in any group identity element is always self inverse so number of elements of order $2$ should be odd(since self inverse elements other than identity are of order $2$ only). Therefore option $1$ is false, for example you can consider $S_3$ group of symmetry on $3$ symbols it has $3$ elements of order $2$. So option $2$ s correct. Option $3$ can be easily discarded using the Syllow theorem.
answered Jan 5 at 16:13
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
497
answered Jan 5 at 16:13
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
497
answered Jan 5 at 16:13
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
497
answered Jan 5 at 16:13
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
497
497
$begingroup$
Suppose $G$ is a group of even order and it does not have any element of order 2. Let $a in G^*$ then o(a)>2 so $a^{-1}$ will be distinct from $a$ and will belong to G. So it can be seen that any element of $G$ other than identity must occur in pair in $G$. So order of $G$ become odd including identity. Which leads to a contradiction. So $G$ will have an element of order 2 so a subgroup of order 2 also.
$endgroup$
– MANI SHANKAR PANDEY
Jan 5 at 17:51
add a comment |
$begingroup$
Suppose $G$ is a group of even order and it does not have any element of order 2. Let $a in G^*$ then o(a)>2 so $a^{-1}$ will be distinct from $a$ and will belong to G. So it can be seen that any element of $G$ other than identity must occur in pair in $G$. So order of $G$ become odd including identity. Which leads to a contradiction. So $G$ will have an element of order 2 so a subgroup of order 2 also.
$endgroup$
– MANI SHANKAR PANDEY
Jan 5 at 17:51
$begingroup$
Suppose $G$ is a group of even order and it does not have any element of order 2. Let $a in G^*$ then o(a)>2 so $a^{-1}$ will be distinct from $a$ and will belong to G. So it can be seen that any element of $G$ other than identity must occur in pair in $G$. So order of $G$ become odd including identity. Which leads to a contradiction. So $G$ will have an element of order 2 so a subgroup of order 2 also.
$endgroup$
– MANI SHANKAR PANDEY
Jan 5 at 17:51
$begingroup$
Suppose $G$ is a group of even order and it does not have any element of order 2. Let $a in G^*$ then o(a)>2 so $a^{-1}$ will be distinct from $a$ and will belong to G. So it can be seen that any element of $G$ other than identity must occur in pair in $G$. So order of $G$ become odd including identity. Which leads to a contradiction. So $G$ will have an element of order 2 so a subgroup of order 2 also.
$endgroup$
– MANI SHANKAR PANDEY
Jan 5 at 17:51
add a comment |
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2
$begingroup$
Whatever $K_4$ is, it isn't every possible group of even order.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 15:33
1
$begingroup$
I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
$endgroup$
– paw88789
Jan 5 at 15:34
$begingroup$
@LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
$endgroup$
– jasmine
Jan 5 at 15:34