# Possible order of a group that has a subgroup of order 7

I need the result of the following question, however I cannot solve it myself. Dear fellows please help me. Thank you in advance.

The statement is:

Know $G$ is a finite group and it has a subgroup $H$ of order 7, and $H$ has no element except the identity is its own inverse, determine the possible order of group $G$.

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Saying that "$H$ has no element except the identity [that] is its own inverse" does not really add information. Such an element would have order 2, and since 2 does not divide 7 there is no way it can exist in a (sub)group of order 7. – Henning Makholm Oct 12 '11 at 20:43

Edit: I misread the question as asking that $G$ have no element other than the identity that is its own inverse. What follows answer under that assumption.

The answer is that the order of $G$ can be any odd multiple of $7$.

That the order of $G$ must be a multiple of $7$ follows from Lagrange's Theorem.

That the order of $G$ cannot be even follows from Cauchy's Theorem (if $2$ divides the order of $G$, then $G$ would have an element of order $2$; i.e., an element other than the identity that is its own inverse). Alternatively, it is a standard exercise to show that if $G$ has even order, then it must have an element other than the identity that is equal to its own inverse: simply define an equivalence relation on $G$ by letting $x\sim y$ if and only if $x=y$ or $x=y^{-1}$; this partitions $G$ into equivalence classes, and each equivalence class has either two elements (if the elements are not equal to their inverses) or one element (if the element equals its own inverse). Let $E$ be the number of classes with two elements, and $O$ the number of classes with two elements. Then $|G| = 2|E|+O$. Since $|G|$ and $2|E|$ are both even, then $|O|$ is even. But $|O|$ is at least $1$, since $e$ is its own inverse, so $|O|\geq 2$, and so there must be some element other than the identity that is equal to its own inverse. Thus, if $|G|$ is even, then $G$ cannot satisfy the condition "no element other than the identity is its own inverse."

That any odd multiple of $7$ can be the order of $G$ follows by taking the cyclic group of order $7k$ with $k$ odd, which satisfies the given conditions.

But the problem says $H$ has no elements other than the identity that are their own inverse. Well, that is no restriction on $H$! No group of odd order (including $H$, which has order $7$) can have an element other than the identity that equals its own inverse: if $x\in H$ is such that $x=x^{-1}$, then $\langle x\rangle = \{x,e\}$, so $|\langle x\rangle|=1$ or $2$; since by Lagrange's Theorem the order of $\langle x\rangle$ must divide $|H|=7$, then $|\langle x\rangle|=1$, hence $x=e$.

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I read the problem as only claiming that $H$ has no element of order 2. – Zev Chonoles Oct 12 '11 at 19:44
The question asks however that $H$ not have an element of order 2, though I suspect the OP intended the question as you wrote it. – SL2 Oct 12 '11 at 19:44
@SL2: oops, quite right. – Arturo Magidin Oct 12 '11 at 19:46
Thanks. The missing part for me is the Cauchy's theorem. – newbie Oct 12 '11 at 19:48
@newbie: It's a very easy exercise that if a finite group has an even number of elements, then there must exist an element other than the identity that is its own inverse; you don't need the full force of Cauchy for it, and chances are you have already proven it during the course. – Arturo Magidin Oct 12 '11 at 19:50

Lagrange's Theorem will be helpful here. It shows that if $G$ has a subgroup $H$ of order 7, then $7=|H|$ must divide $|G|$. So, any such group $G$ will have an order that is divisible by 7.

However, you need to consider whether any number that is divisble by 7 occurs as the order of such a $G$ (perhaps it is possible that even though 21 is divisible by 7, no group $G$ of order 21 has a subgroup of order 7). See if you can get this part on your own.

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