Prove or disprove that there exists no $G$ structure with its bordism group $\Omega_1^{G} =\mathbb{Z}/N$ for $N>2$ It can be found that there are the following bordism group $\Omega_0^{G}$ at $d=0$ and 1 dimensions by requiring $G$ structure for $d$-manifolds:
$$
\Omega_0^{SO} = \mathbb{Z} , \quad \Omega_1^{SO} =  0.
$$
$$
\Omega_0^{Spin} = \mathbb{Z} , \quad \Omega_1^{Spin} =  \mathbb{Z}/2 .
$$
$$
\Omega_0^{O} = \mathbb{Z}/2 , \quad \Omega_1^{O} =  0.
$$
The group classification as the cobordism class is obtained by comparing any two sets of $d$-manifolds $M_1$ and $M_2$ and by seeking whether there are $d+1$-manifolds $\Sigma$ making $M_1$ and $M_2$ as $\Sigma$'s boundary. If yes, then $M_1$ and $M_2$ are in the same cobordism class or correspond to the same group element in the bordism group. All $M_1, M_2, \Sigma$ should have equipped with $G$ structure.
So generally we look at:
$$
\Omega_0^{G}  =  ?  , \quad \Omega_1^{G} =  ?.
$$

*

*The manifold generates $n \in \mathbb{Z}$ class for $d=0$ in $\Omega_0^{SO}=\Omega_0^{Spin} = \mathbb{Z}$ must be an integer $\mathbb{Z}$ number of $d=0$ points (as manifold generators). Is this true that we cannot regard the $n_1,n_2 \in \mathbb{Z}$ points as the boundary of 1d lines, if $n_1 \neq n_2$? But only if $n_1 = n_2=n$, we can use $n$ 1d lines to connect the two sets of $n$ points as boundaries of $n$ 1d lines?


*Since $\Omega_0^{O} = \mathbb{Z}/2$, this means two $d=0$ points can be cobordant to nothing via an unoriented $d=1$ line segment. I may conjecture that:

For any $G$ structure, its bordism group must be either $\Omega_0^{G} =\mathbb{Z}$ or $\Omega_0^{G} =\mathbb{Z}_2$. Namely, it cannot be $\Omega_0^{G} =\mathbb{Z}/N$ for $N\neq 2$ or
or $\Omega_0^{G} =0$  for any $G$.  Is this correct?



*I may conjecture that:


there exists no $G$ structure with its bordism group $\Omega_1^{G} =\mathbb{Z}$. Namely, we only have either
$\Omega_1^{G} =\mathbb{Z}/N$
or
$\Omega_1^{G} =0$ for any $G$. Is this correct?
(Is this true that we must have $\Omega_1^{G} =\mathbb{Z}/N$ with $N =1$ as 0 and $N =2$?)

If so:
This means that for some finite $N$, the $N$ sets of $1$-manifolds ($S^1$) must be regarded as some bundary of a $G$ structure $2$-manifold. Thus, the $N$ sets of $S^1$ must be cobordant to nothing via this $G$ structure $2$-manifold. Can we prove this? Is this true that $N\leq 2$?
 A: $\newcommand{\Z}{\mathbb Z}$Part 2 is correct: for any $G$-structure, $\Omega_0^G$ is either isomorphic to $\Z$ or
$\Z/2$. Part 3 isn't correct, and I'll give a counterexample.
First, part 2: let $\rho\colon G\to O$ be a $G$-structure. $\Omega_0^G$ is isomorphic to $\pi_0(MG)$, where $MG$ is
the Thom spectrum of the pullback of the tautological bundle $V\to BO$ across $B\rho\colon BG\to BO$. Since $MG$ is
connective, the Hurewicz map $\pi_0(MG)\to \tilde H_0(MG)$ is an isomorphism. The Thom isomorphism implies
$\tilde H_0(MG)\cong H_0(BG; \Z_{\sigma})$, where $\Z_\sigma$ is the orientation local system for $B\rho^*V$.
For a general space $X$, the zeroth twisted homology with coefficients in a $\Z[\pi_1(X)]$-module $V$ is the
coinvariants $V/\{v - \gamma\cdot v\mid \gamma\in \pi_1(X)\}$ (See
Davis-Kirk, Proposition 5.14.) In our case, only two
things can happen.

*

*If $B\rho^*V$ is orientable, $\Z_\sigma$ is a trivial local system, and we obtain $H_0(BG;\Z_\sigma) \cong \Z/0 =
  \Z$.

*If $B\rho^*V$ is unorientable, $\Z_\sigma$ is a nontrivial local system. This means there is an element of
$\pi_1(BG)$ which sends $1\mapsto -1$, so we're quotienting by a subgroup containing $2\Z$. However, as
$\pi_1(BG)$ is a group, its action cannot send $1$ to $0$, so we're quotienting out by exactly $2\Z$, and we
obtain $H_0(BG;\Z_\sigma)\cong \Z/2$.

For part 3: for a $G'$-structure with $\Omega_1^{G'}\cong\Z$, consider $G' = \mathrm{SO}\times\Z$. This encodes the notion
of oriented manifolds with a principal $\Z$-bundle, or equivalently oriented manifolds with a map to $S^1$. The
Atiyah-Hirzebruch spectral sequence
$$E_2^{p,q} = H_p(S^1; \Omega_q^{\mathrm{SO}}) \Longrightarrow \Omega_{p+q}^{\mathrm{SO}}(S^1) = \Omega_{p+q}^{G'}$$ collapses in
degrees $3$ and below, producing an isomorphism $\Omega_1^{G'} \cong H_1(S^1)\cong\Z$. One can take as a generator
the $1$-manifold $S^1$ with principal $\Z$-bundle $\mathbb R\to \mathbb R/\Z\cong S^1$.
