# Thread: Groups & Subgroups help

1. ## 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

2. 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

3. 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.

4. 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, $\displaystyle (12)(23)=(132)$).

5. Thanks for the reply.

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.

6. Originally Posted by craig
Thanks for the reply.

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.

7. 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?

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 $\displaystyle D_{12}$?

8. 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

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.

9. Thanks again!