Cohomology group and elliptic curve Let $E$ be an elliptic curve with a 3-torsion point $P$ and $G = \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Let $X = \{O, P, -P\}$ where $O$ is the point at infinity and $X$ is a $G$-module. Why does $H^{1}(G, X)$ inject into $\operatorname{Sel}_{3}(E/\mathbb{Q})$?
 A: It doesn't (always)!
Nor would you expect it to. Assume $X$ is $G_{\mathbb{Q}}$ stable. Then in general, the group $H^1(\mathbb{Q},X)$ will be infinite whereas the selmer group $Sel_{3}(E/\mathbb{Q})$ finite.
Even more it may not even be the case $H^1(\mathbb{Q},X)$ injects into $H^1(\mathbb{Q},E[3]).$ Here's an exmple:
Consider the curve $E$ given by $y^2 +xy + y = x^3 -171x -874.$ One checks that  $E[3]$ has three $\mathbb{Q}(\mu_3)$-points and one $\mathbb{Q}$-point. Therefore, if you let $X$ be the subgroup of $E[3]$ consisting of these $\mathbb{Q}(\mu_3)$-points, then $X$ is Galois stable and $G_\mathbb{Q}$ acts on $X$ through the nontrivial character of $Gal(\mathbb{Q(\mu_3)}/\mathbb{Q}).$ It follows that $X \cong \mu_3$ as a Galois module. 
Because $E[3]$ is reducible with a stable one dimensional subspace $X,$ we have $E[3]$ is an extension of a character $\chi$ by $X \cong \mu_3.$ By the weil pairing $\chi \otimes X = \chi \otimes \mu_3 = \mu_3$ and hence $\chi$ is the trivial character. 
From the short exact sequence $$0\rightarrow X \rightarrow E[3] \rightarrow \chi\rightarrow 0,$$ one obtains the exact sequence 
$$E[3](\mathbb{Q}) \rightarrow \mathbb{F}_3 \rightarrow H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3]).$$ 
And since $E[3](\mathbb{Q}) = 0,$ it follows that the map $H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3])$ is not injective. 
(Conversely, observe that if $H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3])$ is not injective then $E[3]$ must be a nonsplit extension of the trivial character by $\mu_3.)$
The problem is the incongruity in your question. You have assumed lots of local conditions on $Sel_3(E/\mathbb{Q})$ but none on $H^1(\mathbb{Q},X).$ What are the local conditions one might consider for cocylces in $H^1(\mathbb{Q},X)?$
Well, recall that we can attach a selmer group to any $\mathbb{Q}$-isogeny between elliptic curves. For example $Sel_3(E/\mathbb{Q})$ is the selmer group attached to the multiplication by $3$-isogeny. There is a natural isogeny attached to $X,$ namely the projection map $\psi_X: E \rightarrow E/X,$ and because $X$ is Galois stable $\psi_X$ is defined over $\mathbb{Q}.$ We define the Selmer group attached to $\psi := \psi_X$ as the kernel 
$$Sel^{\psi}(E/\mathbb{Q}) := \ker(H^1(\mathbb{Q},X) \rightarrow \prod_p H^1( \mathbb{Q}_p,E)).$$
Because multiplication by $3$ factors through $X,$ (or rather because $X \subset E[3]$), we have a map $ Sel^{\psi}(E/\mathbb{Q}) \rightarrow Sel_3(E/\mathbb{Q})$ induced by the map $X \rightarrow E[3].$ One easily checks that the map
$$H^1(\mathbb{Q},X) \rightarrow \prod_p H^1( \mathbb{Q}_p,E)$$
factors through
$$H^1(\mathbb{Q},E[3]) \rightarrow \prod_p H^1( \mathbb{Q}_p,E)$$
and hence $ Sel^{\psi}(E/\mathbb{Q}) \rightarrow Sel_3(E/\mathbb{Q})$ is injective whenever $H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3])$ is injective. 
What about when $H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3])$ isn't injective i.e. when  $X \cong \mu_3$ and $E[3]$ is a nonsplit extension of the trivial representation by $X?$ Here one sees using the argument above that the map between selmer groups still has a 1-dimensional kernel. Oh well.   
