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As given definition of group action goes like this :

For any group G and X is a set then $\phi : G \times X \to X$ and such that $\phi(1,x)=x$ and $\phi(g,\phi(h,x))=\phi(gh,x) $ for every $x \in X$ and every $g,h \in G$

Does this imply that X is subset of G?If not why is it so?

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    $\begingroup$ Why do you think it does? As a dumb example, say $G = \{e\}$ and $X$ any set with more than one element — there's no chance of even finding an injection $X \hookrightarrow G$. $\endgroup$ – Hoot Nov 11 '14 at 17:56
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    $\begingroup$ $X$ does not have to be a subset of $G$; this is a common point of confusion with group actions. An equivalent definition of a group $G$ action on a set $X$ is a homomorphism $\phi: G \rightarrow S_{X}$, the set of permutations on $X$. Tim Gowers has a very nice post on this: gowers.wordpress.com/2011/11/06/group-actions-i $\endgroup$ – Alex Wertheim Nov 11 '14 at 17:58
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Not usually. The set $X$ can have $G$ as a subset, and $G$ can certainly act on itself, but this is not what you're thinking! The best examples that I can think of are the dihedral groups. Take your favorite one, $D_n$, and let the set be the vertices of a $n$-gon. Then $G$ acts on the vertices via group action, but $G$ is most certainly not in the set.

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