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The pentagon has 5 line symmetries and therefore we will have 10 symmetries. So, we let the group G with order 10 denote the symmetry group of a pentagon.

A subset $H$ of $G$ is a subgroup $(H, *)$ to the group $(G, *)$ if and only if $(H,*)$ is a group. To determine how many subgroups there are, I can use the Lagrange's theorem which tells us that the subgroups of a group with order n have a order m such that $m | n$.

By this theorem we get know that the group $G$ has 5 subgroups since the divisors m of 10 is $m=1,2,5,10$.

The question is, how shall I sketch the lattice of subgroups? Shall I check every possible subset and then check if they are subgroups or is there a faster way?

I know that the 10 symmetries are the identity element, 4 rotation and 5 reflections. We can see these operation of the transformation as permutation.

The first transformation that rotates 72 degrees are the permutation (ABCDE), 144 degrees are the permutation (ABCDE)^2 etc. This permutation have the order 5 since if we rotate 5 times we will get the identity ("the original pentagon") .

The reflection transformations are permutations with 2 cycles with length 2.

But how do I know that the subgroups with order 5 are the ones with identity and 4 rotations?

EDIT

The order 2 is easy to determine. The subsets $\{i,g\}$ there $g \in G -\{reflections\}$ are not groups because the inverse of g is not in the subset. BUT if g is one of the reflections then we have a group, since the reflections have the order 2 which means that the element $g$ is the inverse of it self. This means that the subset is a subgroup because (i) it is closed and (ii) the inverse is in the subset.

EDIT 2

Can I think like this? We let a subset be $$H = \{\text{id},r,r^2,r^3,z\}$$ there r are the rotation and z is the reflection. By checking the properties for a group, we can see that the properties about that every element in H has an inverse does not hold. the element $r$ has the inverse $r^4$ which is not in $H$ and therefore the subset $H$ cannot be a subgroup. So, by removing one of the rotations and putting one of the reflections, we see that we are removing some inverses. So, the only subgroup of order 5 are $$K=\{\text{id},r,r^2,r^3,r^4\}$$

Am I thinking correctly? Or is it better to write down the group table to find the subsets which are subgroups?

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  • $\begingroup$ The reflections all have order two. So by Lagrange's theorem none of them can be contained in a subgroup of order five. $\endgroup$ – Jyrki Lahtonen May 23 '14 at 19:34
  • $\begingroup$ I am abit confused over the group order and element order. Group order means the number of element in a group and element order means the least positiv integer such that $x^n=1$ right? $\endgroup$ – Angelica May 23 '14 at 19:40
  • $\begingroup$ That's correct. And Lagrange's theorem tells (among other things) that the order of an element of a finite group is always a factor of the order of the whole group. $\endgroup$ – Jyrki Lahtonen May 23 '14 at 19:41
  • $\begingroup$ Note that the order of an element is (almost by definition) the order of the cyclic subgroup generated by it. $\endgroup$ – Nishant May 23 '14 at 19:42
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    $\begingroup$ Slightly relevant nit pickery: you can't in principle count the number of subgroups of a group just by counting the number of divisors of its order, because a group can have many different subgroups (even non-isomorphic subgroups) of the same order. In this case, you know that any subgroup of $G$ order $5$ must be a $C_5$ because $C_5$ is the only group of order $5$, but that won't always hold. $\endgroup$ – Steven Stadnicki May 23 '14 at 22:24
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Your thinking is more or less correct. Generally, when we wish to think about the subgroups of a group $G$, we take different subsets of $G$ and look at what groups those subsets generate.

The obviously thing to do is start with one element. Let's say I start with a rotation $r^k$ for some $k = 1, 2, 3, 4$. Since $5$ is a prime, the numbers $1, 2, 3, 4$ have multiplicative inverses mod $5$. This implies that $\langle r^k\rangle$ actually contains $r$. Hence, $\langle r^k\rangle = \langle r\rangle = \{1, r, r^2, r^3, r^4\}$ for each $k$. Now, let's say I start with a reflection $z$ instead. It's easy to see that $\langle z\rangle = \{1, z\}$. Thus, we have $5$ different subgroups of order $2$, one for each reflection. The only remaining case is $\langle 1\rangle = \{1\}$.

Now let's consider a generating set with two elements. If we take $\langle r^k, r^l\rangle$ for some $k, l$, this is clearly just $\langle r\rangle$ again. Suppose instead we take $\langle r^k, z\rangle$. Then this subgroup contains $r$ and $z$, hence it contains $r^l z$ for each $l=0, 1, 2, 3, 4$, i.e. it contains every reflection. It also contains the powers of $r$, which are the rotations, hence $\langle r^k, z\rangle = G$. Finally, we could take $\langle z_1, z_2\rangle$ for two reflections $z_1, z_2$. But then $z_1 z_2$ is a rotation which is in $\langle z_1, z_2\rangle$, hence by the previous case we have $\langle z_1, z_2\rangle = G$.

From here it's easy to see that we've found every subgroup of $G$. Obviously this took a lot less time than checking all $2^{10}$ subsets of $G$.

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  • $\begingroup$ I did understand everything until $\langle r^k, r^l\rangle$. I have never seen a generator as that before. Can 2 generator generate a group or what does it mean :)? and thanks for a detailed answer. $\endgroup$ – Angelica May 23 '14 at 21:55
  • $\begingroup$ Yes. In general, if $S$ is a subset of $G$, by $\langle S \rangle$ we mean the smallest subgroup of $G$ containing $S$. This is known as the subgroup of $G$ generated by $S$. $\endgroup$ – Alex G. May 23 '14 at 23:25
  • $\begingroup$ Could you link some site who has some information about $\langle r^k, r^l\rangle$ ? I did look into my book but there were nothing about the generating set with two elements. $\endgroup$ – Angelica May 23 '14 at 23:28
  • $\begingroup$ Here's the wikipedia page: en.wikipedia.org/wiki/Generating_set_of_a_group $\endgroup$ – Alex G. May 23 '14 at 23:29
  • $\begingroup$ I'm not sure what you're asking. Could you rephrase? $\endgroup$ – Alex G. May 23 '14 at 23:34

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