subgroups of the group of pentagon symmetries 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?
 A: 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 $G$.
