# Groups & Subgroups help

• May 2nd 2010, 10:48 AM
craig
Groups & Subgroups help
Got a tricky question here that I'm not too sure how to approach:

$\displaystyle F$ is a group of order 12 under multiplication and with $\displaystyle e$ the neutral element.

i) $\displaystyle H = \{ f \in F : f^{-1} = f \}$ is the set of self inverse elements of $\displaystyle F$.

I need to show that $\displaystyle H$ is a subgroup of $\displaystyle F$ if $\displaystyle F$ is abelian.

ii) We now need to assume that $\displaystyle F$ is cyclic, with generator $\displaystyle \alpha$ and find the order and all the elements of $\displaystyle H$

iii) Give an example of a group $\displaystyle F$ of order 12 for which $\displaystyle H = \{ f \in F : f^{-1} = f \}$ isn't a subgroup. (For this one obviously you need to use the sub-group criteria, just not sure how to apply them).

Greatly appreciate any help/pointers with this.

Craig
• May 2nd 2010, 12:06 PM
rubic
i) $\displaystyle H = \{ f \in F : f^{-1} = f \}$ is the set of self inverse elements of $\displaystyle F$.

I need to show that $\displaystyle H$ is a subgroup of $\displaystyle F$ if $\displaystyle F$ is abelian.

SG1 = Since neutral element is its self inverse, H is not empty

SG2 = a*b member of H, for all a,b member of H
$\displaystyle (a*b)^{-1}= (b^{-1})*(a^{-1})= b*a$
since f is abelian $\displaystyle b*a=a*b$
$\displaystyle (a*b)^{-1}= a*b$
Hence it satisfies conditon of H
SG2 holds
SG3 = every element is its self inverse

$\displaystyle a^{-1}=a,$
Thus $\displaystyle a^{-1}$ member of H
• May 3rd 2010, 12:29 AM
craig
Thanks for the reply, it actually sounds quite simple when you put it like that ;)

For the part ii). I need to try and find a way to list the elements of $\displaystyle F$ in terms of $\displaystyle \alpha$, how would I go about this, something like $\displaystyle x (mod \alpha)$ or would it involve powers of $\displaystyle \alpha$, considering the group is under multiplication?

And for the third section, obviously need to find a group that's not a subgroup. Seeing as SG1 and 3 hold, need to find a non-abelian group with order 12. Would the Dihedral Group of order 12, $\displaystyle D_{12}$ be such a group. If so how would I list the elements, in terms of $\displaystyle rot_{\beta}$ where $\displaystyle \beta$ are the 12 degrees of rotation?

Thanks again for the help, really appreciate it.
• May 3rd 2010, 01:40 AM
Swlabr
Quote:

Originally Posted by craig
Got a tricky question here that I'm not too sure how to approach:

$\displaystyle F$ is a group of order 12 under multiplication and with $\displaystyle e$ the neutral element.

i) $\displaystyle H = \{ f \in F : f^{-1} = f \}$ is the set of self inverse elements of $\displaystyle F$.

I need to show that $\displaystyle H$ is a subgroup of $\displaystyle F$ if $\displaystyle F$ is abelian.

ii) We now need to assume that $\displaystyle F$ is cyclic, with generator $\displaystyle \alpha$ and find the order and all the elements of $\displaystyle H$

iii) Give an example of a group $\displaystyle F$ of order 12 for which $\displaystyle H = \{ f \in F : f^{-1} = f \}$ isn't a subgroup. (For this one obviously you need to use the sub-group criteria, just not sure how to apply them).

Greatly appreciate any help/pointers with this.

Craig

For ii) look at the cyclic group of order 12. This, well, is the only cyclic group of order 12...!

So, for which n does $\displaystyle (\alpha^n)^2 = 1$ hold?

For ii) you clearly have that any subgroup is closed under inverses. So, you need to look at either your favourite or second-favourite non-abelian group of order 12 (it definately doesn't hold for your third favourite non-abelian group of order 12, T) and find two elements in it of order 2 which, when multiplied together, give you an element not of order 2.

I would suggest looking in $\displaystyle D_{12}$, the dihedral group of order 12 (it contains a copy of $\displaystyle S_3$ in it, and $\displaystyle S_3$ is generated by its 2-cycles. For example, $\displaystyle (12)(23)=(132)$).
• May 3rd 2010, 05:11 AM
craig

Quote:

Originally Posted by Swlabr
For ii) look at the cyclic group of order 12. This, well, is the only cyclic group of order 12...!

So, for which n does $\displaystyle (\alpha^n)^2 = 1$ hold?

With it being cyclic, you have the following.

$\displaystyle \{ e \alpha, \alpha^2, \alpha^3 ... \alpha^{11} \}$

$\displaystyle \{ e, \alpha^2, \alpha^4 ... \alpha^{10} \}$

... etc

$\displaystyle \{ e = \alpha^{12} \}$

So the only element of $\displaystyle H$ would be $\displaystyle \alpha^{12} = \langle e \rangle$, which has order 1?

For part 3, I considered $\displaystyle D_{12}$, the symmetries of a regular hexagon.

For this would I just list the 6 symmetries and the 6 reflections in $\displaystyle D_{12}$, and to show that it is not a subgroup explain how it's not commutative?

Thanks again for the pointers.
• May 3rd 2010, 07:09 AM
Swlabr
Quote:

Originally Posted by craig

With it being cyclic, you have the following.

$\displaystyle \{ e \alpha, \alpha^2, \alpha^3 ... \alpha^{11} \}$

$\displaystyle \{ e, \alpha^2, \alpha^4 ... \alpha^{10} \}$

... etc

$\displaystyle \{ e = \alpha^{12} \}$

So the only element of $\displaystyle H$ would be $\displaystyle \alpha^{12} = \langle e \rangle$, which has order 1?

For part 3, I considered $\displaystyle D_{12}$, the symmetries of a regular hexagon.

For this would I just list the 6 symmetries and the 6 reflections in $\displaystyle D_{12}$, and to show that it is not a subgroup explain how it's not commutative?

Thanks again for the pointers.

In $\displaystyle \mathbb{Z}_{12}$ there is one element of order 2. In fact, $\displaystyle \mathbb{Z}_{n}$ has a subgroup of order $\displaystyle m$ if and only if $\displaystyle m \mid n$.

Write down all the elements of $\displaystyle \mathbb{Z}_{12}$ in a list, and beside them write down their squares. This will reveal the element of order 2.

For $\displaystyle D_{12}$, again list all the elements and their squares. This will give you the elements of order 2. Then, take two elements of order 2 and multiply them together. Do this until you find a pair which does not appear in your list of elements of order 2.
• May 3rd 2010, 07:35 AM
craig
Quote:

Originally Posted by Swlabr
In $\displaystyle \mathbb{Z}_{12}$ there is one element of order 2. In fact, $\displaystyle \mathbb{Z}_{n}$ has a subgroup of order $\displaystyle m$ if and only if $\displaystyle m \mid n$.

Write down all the elements of $\displaystyle \mathbb{Z}_{12}$ in a list, and beside them write down their squares. This will reveal the element of order 2.

Ahh thankyou, the only element of order 2 in $\displaystyle \mathbb{Z}_{12}$ would be 6 right, because this cycle is $\displaystyle (0,6)$?

Would this mean that 6 is it's own inverse in $\displaystyle \mathbb{Z}_{12}$ then?

Quote:

Originally Posted by Swlabr
For $\displaystyle D_{12}$, again list all the elements and their squares. This will give you the elements of order 2. Then, take two elements of order 2 and multiply them together. Do this until you find a pair which does not appear in your list of elements of order 2.

When you say the elements, do you mean the different rotations and reflections in $\displaystyle D_{12}$?
• May 3rd 2010, 08:05 AM
Swlabr
Quote:

Originally Posted by craig
Ahh thankyou, the only element of order 2 in $\displaystyle \mathbb{Z}_{12}$ would be 6 right, because this cycle is $\displaystyle (0,6)$?

Would this mean that 6 is it's own inverse in $\displaystyle \mathbb{Z}_{12}$ then?

yup

Quote:

Originally Posted by craig
When you say the elements, do you mean the different rotations and reflections in $\displaystyle D_{12}$?

Yes - the elements of the group are the things generated by the rotations and reflections.
• May 3rd 2010, 08:06 AM
craig
Thanks again!