Let $G$ be a finite group of even order. Then which of the following statements is correct?












1















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







share|cite|improve this question




















  • 2




    Whatever $K_4$ is, it isn't every possible group of even order.
    – Lord Shark the Unknown
    2 days ago






  • 1




    I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
    – paw88789
    2 days ago












  • @LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
    – jasmine
    2 days ago


















1















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







share|cite|improve this question




















  • 2




    Whatever $K_4$ is, it isn't every possible group of even order.
    – Lord Shark the Unknown
    2 days ago






  • 1




    I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
    – paw88789
    2 days ago












  • @LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
    – jasmine
    2 days ago
















1












1








1


0






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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Chinnapparaj R

5,3131827




5,3131827










asked 2 days ago









jasminejasmine

1,599416




1,599416








  • 2




    Whatever $K_4$ is, it isn't every possible group of even order.
    – Lord Shark the Unknown
    2 days ago






  • 1




    I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
    – paw88789
    2 days ago












  • @LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
    – jasmine
    2 days ago
















  • 2




    Whatever $K_4$ is, it isn't every possible group of even order.
    – Lord Shark the Unknown
    2 days ago






  • 1




    I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
    – paw88789
    2 days ago












  • @LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
    – jasmine
    2 days ago










2




2




Whatever $K_4$ is, it isn't every possible group of even order.
– Lord Shark the Unknown
2 days ago




Whatever $K_4$ is, it isn't every possible group of even order.
– Lord Shark the Unknown
2 days ago




1




1




I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
– paw88789
2 days ago






I'm guessing $K_4$ is the Klein 4-group, which has three elements of order 2.
– paw88789
2 days ago














@LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
– jasmine
2 days ago






@LordSharktheUnknown $mathbb{Z}_{2n}$ for $n=1,2,3.......$
– jasmine
2 days ago












2 Answers
2






active

oldest

votes


















4














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!






share|cite|improve this answer























  • im not getting why u write $ |G|= 1 + k $?
    – jasmine
    2 days ago






  • 1




    That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
    – Chinnapparaj R
    2 days ago



















1














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.






share|cite|improve this answer





















  • 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.
    – MANI SHANKAR PANDEY
    2 days ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














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!






share|cite|improve this answer























  • im not getting why u write $ |G|= 1 + k $?
    – jasmine
    2 days ago






  • 1




    That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
    – Chinnapparaj R
    2 days ago
















4














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!






share|cite|improve this answer























  • im not getting why u write $ |G|= 1 + k $?
    – jasmine
    2 days ago






  • 1




    That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
    – Chinnapparaj R
    2 days ago














4












4








4






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!






share|cite|improve this answer














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!







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Chinnapparaj RChinnapparaj R

5,3131827




5,3131827












  • im not getting why u write $ |G|= 1 + k $?
    – jasmine
    2 days ago






  • 1




    That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
    – Chinnapparaj R
    2 days ago


















  • im not getting why u write $ |G|= 1 + k $?
    – jasmine
    2 days ago






  • 1




    That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
    – Chinnapparaj R
    2 days ago
















im not getting why u write $ |G|= 1 + k $?
– jasmine
2 days ago




im not getting why u write $ |G|= 1 + k $?
– jasmine
2 days ago




1




1




That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
– Chinnapparaj R
2 days ago




That means $$vert G vert=vert {e} vert + vert {a in G :a neq a^{-1} } vert +text{number of remaining order $2$ elements} $$
– Chinnapparaj R
2 days ago











1














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.






share|cite|improve this answer





















  • 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.
    – MANI SHANKAR PANDEY
    2 days ago
















1














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.






share|cite|improve this answer





















  • 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.
    – MANI SHANKAR PANDEY
    2 days ago














1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









MANI SHANKAR PANDEYMANI SHANKAR PANDEY

477




477












  • 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.
    – MANI SHANKAR PANDEY
    2 days ago


















  • 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.
    – MANI SHANKAR PANDEY
    2 days ago
















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.
– MANI SHANKAR PANDEY
2 days ago




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.
– MANI SHANKAR PANDEY
2 days ago


















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