Is a subgroup of a fundamental group a fundamental group? Let $(X,\ast)$ be a based topological space (maybe path connected or not, I don't know if this will be relevant to the solution).
Let $\pi:=\pi_1(X,\ast)$ be its fundamental group and let $H$ be any nontrivial subgroup of $\pi_1$ (suppose that there exists one).
My question is: is there some subspace $Y\subset X$ such that $H=\pi_1(Y,\ast)$?
In resume: given a subgroup of a fundamental group, is it also a fundamental group of some subspace of the total space?
 A: I'm interpreting your question as follows:  when you write $H = \pi_1(Y,\ast)$, you mean the inclusion $i:Y\rightarrow X$ induces an injective map on $\pi_1$ with image $H$.  (As opposed to $\pi_1(Y)$ being abstractly isomorphic to $H$).
With this interpretation, the answer is no.  For example, consider $X = S^1$.  It is well known that $\pi_1(X) = \mathbb{Z}$.  Now, consider the subgroup $H = 2\mathbb{Z}$.  I claim that there is no proper subspace $Y$ for which which has this fundamental group.
We may assume wlog that $Y$ is connected.  Then note that any connected proper subset of $S^1$ is homeomorphic to a connected subset of $(0,1)$.  These are easy to classify, they are, up to homeomorphism, $(0,1)$, $[0,1]$, and $(0,1]$.  None of these has infinite cyclic fundamental group.
Edit  Here is an example with $H$ not even abstractly isomorphic to a particular subgroup of $\pi_1(X)$.
Take $X = S^1 \vee S^1$, the wedge sum of 2 $S^1$s.  The fundamental group of $X$ is known to be isomorphic to the free group on two generators.  It's also know that a free group on two generators contains subgroups isomorphic to the free group on $n$ generators for any finite $n$ (we make even take $n$ to be countable).  Further, it's know that free groups on different numbers of generators are never isomorphic.
Let $H$ denote any of these subgroups for $n > 2$.  In particular, $H$ is not isomorphic to either $0$, $\mathbb{Z}$, or the free group on two generators.  I claim that no subspace $Y$ of $X$ has $H$ as a fundamental group, even up to abstract isomorphism.
As above, we may assume $Y$ is connected.  If $Y$ does not contain the wedge point, then it must be contained in a proper portion of one of the two circles, so the above argument shows $\pi_1(Y)$ is trivial.  Hence, $Y$ must contain the wedge point.  Now, if $Y$ does not contain the whole of the first circle, then $Y$ deformation retracts onto a subspace of the other circle, hence, by the previous argument has $\pi_1 = 0$ or $\mathbb{Z}$.  Thus, $Y$ must contain the while first circle.  Likewise, $Y$ must contain the whole second circle.
But then this implies $Y = X$, so $Y$s fundamental group is isomorphic to the free group on $2$ generators, so not isomorphic to $H$.
Final (?) Edit  Here's one in the finite fundamental group case.  Let $X$ be obtained from $S^1$ by attaching a $D^2$ by a degree $4$ map.  A simple van Kampen argument shows $\pi_1(X) = \mathbb{Z}/4$.  Let $H$ be the unique subgroup of $\pi_1(X)$ isomorphic to $\mathbb{Z}/2$.
I claim that no subspace $Y$ has fundamental group abstractly isomorphic to $H$.  If $Y$ misses a point of the interior of the $D^2$, then $Y$ deformation retracts onto a subspace of $S^1$, so by the above argument, has $\pi_1 = 0$ or $\mathbb{Z}$.   Hence, we may assume wlog that $Y$ contains all of the interior of $D^2$.  Likewise, if $Y$ misses a point of $S^1$, then restricting the van Kampen argument to $Y$ shows that $\pi_1(Y) = 0$, so $Y$ must contain all of $S^1$.  This implies $Y = X$, so $\pi_1(Y)\neq H$.
A: Take the space $X:=\mathbb{S}^1$ then $\pi_1(X)=\mathbb{Z}$ But any proper subspace of $X$ has has trivial fundamental group, so no subspace has fundamental group $\mathbb{2Z}\leq\mathbb{Z}$.
