To show that H,K are sub-groups of G (U36) Let $G\:=U_{36}$ and
$$H=\left\{\left[x\right]\in U_{36}\::\:x\equiv 1\pmod{4}\right\},$$
$$K=\left\{\left[y\right]\in U_{36}\::\:y\equiv 1\pmod{9}\right\}.$$
I need to show that $H$ and $K$ are subgroups of $G$.
First of all, lets write them explicitly for clearly view, what we to show finally, so: 
$$G=\left\{\left[1\right],\left[5\right],\left[7\right],\left[11\right],\left[13\right],\left[17\right],\left[19\right],\left[23\right],\left[25\right],\left[29\right],\left[31\right],\left[33\right],\left[35\right]\right\},$$
$$H=\left\{\left[1\right],\left[5\right],\left[13\right],\left[17\right],\left[25\right],\left[29\right],\left[33\right]\right\},$$
$$K=\left\{\left[1\right],\left[19\right]\right\}.$$
Let's start from $H$. It is clearly a subset, so I'll check all the conditions for a group, and if it's a group and sub-set, it is a subgroup of $G$:
Clearly, it isn't empty and $G$ is a group, therefore all that elements are assosiative, because they from the $G$ group.
I claim that $[1]$ is the identity element, because for each $x\in H$ we have $x\equiv 1\pmod{4}$, so
$$x\cdot 1\equiv 1\cdot x\equiv x\pmod{4}.$$
Not correct from here:
And now I want to show that for each $x\in H$ there exists $x^{-1}\in H$ such that $xx^{-1}=\left[1\right]$.
But from modularic arithmetic rules, for each $x_1,x_2\in H$ we have $x_1\equiv 1\pmod{4}$ and $x_2\equiv 1\pmod{4}$, so
$$x_1\cdot x_2\equiv 1\cdot 1\equiv 1\pmod{4}.$$
So as I understand each elements in $H$ is the opposite of each other. But that makes me think that every element is either the identity element for each other, because: 
$$17\cdot 29\equiv 17\equiv 1\pmod{4}.$$
Am I just confused in this or it is a special characteristics of modulo $1$? 
Otherwise, maybe I don't understand how do we define here the identity element and the inverse, because the operator here is modulo, but the identity element, I think about him as multiply, either the inverse. 
Until here 
Otherwise, maybe $1$ is the identity element, but using Euclid algorithm to find the opposite in this cases, randomly I found that for $[5]$ the opposite is $[-7]$, what modulo 36 means $[29]$ ,for [13] it is $[-11]$, Therefore $[25]$ and $[17]$ is the opposite of himself. 
So what about $[1]$, does it have opposite? 
And is it true for from modularic arithmetic rules, that I always have opposites in such examples of U, or I have to check for each element in hand mode? 
For example in K, as I understand now, the opposite of $19$ is itself, the identity element is $1$ (why ? can some one define for me identity in modulo explicitly and explain it to me?), associativity from $G$, and closure also from modularic arithmetic rule, because multiply of $2$ elements modulo, is an element is modulo, right? 
Correct me whenever I didn't explain good enough or missed something please.
 A: I'll assume you mean $G$ is a cyclic group of order $36$. By the looks of it you are considering $\Bbb{Z}/36\Bbb{Z}$. Do note that $33$ is not coprime to $36$, so $[33]\notin G$.
To check that $H$ is a subgroup of $G$ we need to check three conditions:


*

*That $H$ is a non-empty subset of $G$.

*That $H$ is closed under the group operation on $G$.

*That for every $h\in H$ we have $h^{-1}\in H$.


The first part is obvious. The second part you have already verified: If $h_1,h_2\in H$ then 
$$h_1\equiv1\pmod4\qquad\text{ and }\qquad h_2\equiv1\pmod4,$$
 from which it follows that $h_1h_2\equiv1\pmod4$ and therefore $h_1h_2\in H$.
The third condition can be more difficult. In this case it suffices to check that
$$[5]\cdot[29]=[1],\qquad[13]\cdot[25]=[1],\qquad [17]\cdot[17]=[1],$$
and of course $[1]\cdot[1]=[1]$. It follows that $H$ is a group under the group operation on $G$ (restricted to $H$), which means $H$ is a subgroup of $G$.
You can do the same for $K$.

Note that you do not need to check the third condition if your (candidate) subgroup is finite. If a finite subset $H$ of a group $G$ is closed under the group operation on $G$, then for every $h\in H$ there exist distinct positive integers $m$ and $n$ such that $h^m=h^n$. Without loss of generality $m>n$, and we find
$$h^{m-n}=h^m\cdot h^{-n}=h^n\cdot h^{-n}=e,$$
where $e$ denotes the identity element. Then $h^{m-n-1}=h^{-1}$, which shows that $h^{-1}\in H$ because $H$ is closed under the group operation.
