Do the relations in a group carry forward to a subgroup? If $G=\langle\, S\,|\,R\, \rangle$ is a finitely presented group and $H=\langle\, S'\,|\,R'\,\rangle$ is a group with generating set $S'\subseteq G$ then does that make $H$ a subgroup of $G$? 
Also, If $H$ is a subgroup of $G$ then do the generators of $H$ satisfy the relations of $G$ as well? Let me give an example :
Let $G=\langle\, s_i:1\leq i\leq 4\,|\, s_i^2=1 \text{ for all }i \text{ and } (s_1s_2)^2=(s_1s_4)^2=(s_2s_3)^2=(s_3s_4)^2=1\rangle$
Let $H=\langle\, a,b\,|\,ab^{-1}a^{-1}b^{-1}=1\,\rangle$ where $a=s_3s_1$ and $b=s_4s_2$.


*

*Is $H$ a subgroup of $G$ above?

*If so, can we say $ab=s_3s_1s_4s_2=s_3s_4s_1s_2=s_4s_3s_2s_1=s_4s_2s_3s_1=ba$. This means that $H$ should also have the relation $aba^{-1}b^{-1}=1$? This seems wrong to me.


Could someone please clarify what is wrong? 
Thank you.
 A: When forming the group $H=\langle S'|R'\rangle$, you only consider $S'$ as a set. In particular, you forget that $S'$ is a subset of a group, and thus that there are some operations you can do with its elements. So $H$ has no reason in general to be a subset of $G$.
So in your example, when you consider the group $H=\langle a,b |ab^{-1}a^{-1}b^{-1}\rangle $, you don't actually care about who $a$ and $b$ are, so $H$ has no relation with $G$.
On the other hand, if $H$ is really a subgroup of $G$, then any relation that holds in $G$ will of course still hold in $H$.
A: There are two notions of relations in groups: one is defining relations, and other is called laws that group can satisfy. 
Every group $G$ can be written in form $F(X)/N$, where $F(X)$ is a free group on $X$ — some generating set for $G$, and $N$ is the kernel of surjection $\pi: F(X) \to G$. Defining relations $R$ for given $\pi$ is some subset of $N$ such that $N$ is the minimal normal subgroup of $F(X)$ containing $R$, called the normal closure of $R$ in $F(X)$ They are not determined uniquely: from $R$ you can produce $R'$ changing one of its elements to its inverse or multiplying one by another. Their normal closures, of course, will be same.
Laws in group $G$ (more generally, in some class of groups) are different thing: they are words in (infinite) alphabet (alternatively, elements of free group with countable number of generators) with the property that for every homomorphism to $G$ (to every group in class respectively) they fall into its kernel. Typical examples are


*

*$[x, y] := x^{-1}y^{-1}xy$: group satisfying this law is abelian

*$[xy, z][yz, x][zx, y]$: this law holds in every group (i. e. reduced word corresponding to it is empty)

*$[[x, y], z][[y, z], x][[z, x], y]$: this law holds in groups with abelian commutator subgroup


As it follows from definition, if a law holds in $G$, then it holds in every factorgroup and subgroup of $G$. Somewhat less trivial result is that if a law
holds in $G$ and $H$, then in holds in $G \times H$. Conversely, Birkhoff theorem states that every collection of groups closed under passing to subgroups, factors and products is defined by some set of laws.
You can look at two groups: one is $\langle x, y \, | \, x^2, y^2 \rangle$, the infinite dihedral group, and second is $\langle x, y \, | \, g^2 = 1 \,$for every$\, g \rangle$, which is isomorphic to $\mathbb Z / 2 \times \mathbb Z / 2$. Contemplating a bit on this may help.
