Argument for subgroup of a group 
For a fixed element $a$ of a group $G$, prove or disprove that the set $H =\{xa|x \in G\}$
  is a subgroup of $G$.

So one argument I proved it using regular properties, but I was thinking in this case this could be done as well by proving $H = G$ but I don't know how to prove it in this case I mean clearly $H$ is subset of $G$. 
Now I need to prove that also $G$ is subset of $H$. So if I pick element $(xa)\cdot a^{-1}\cdot a \in G$ which is subset of $H$ so this proves that $H = G$, right? 
IF there is any improvement to my argument please let me know.
 A: Hint:
$$\text{For all}\;\;g\in G\;,\;\;g=\left(ga^{-1}\right)a\in H$$
A: you can even argue that there is a bijection from $H$ to $G$ given by the function $\phi(x) = xa$ , you will have to show that this mapping is onto and one-to-one and then this will conclude that $H = G$, hence $H$ is a subgroup because every group is a subgroup of itself
Another method: 
let $g \in H$ so there exists $x \in G$ such that $g=xa$ now, $g^{-1} = (xa)^{-1}=a^{-1}x^{-1}$ and moreover, we know that $a^{-1}x^{-1}a^{-1} \in G$ and so $g^{-1}$ can be written as $g^{-1} = (a^{-1}x^{-1}a^{-1})a$ and so $g \in H$
A: This isn't quite right. You've started with $xa$ and shown $xa\in H$. You need to start with $x\in G$ and write it in the form $ya$ for some $y$. You should, and in fact must, take $y=xa^{-1}$, which is probably the argument you were reaching towards.
A: Ther are lots of good answer for this questions and they have very good explanations. But I think you can just apply the subgroup test.  
Note that $1=(a^{-1})a\in H\not= \emptyset$
Let $g_1=xa$ and $g_2=ya$ be two elements of $H.$ Then $$g_1g_2^{-1}=xaa^{-1}y^{-1}=(xy^{-1}a^{-1})a\in H$$ Hence $H\le G.$
