Should not we eliminate $\pm 1$ from $\mathbb{Z}^\times_N$ in RSA accumulators?

RSA accumulator is an authenticated set built from cryptographic assumptions in hidden-order groups such as $$\mathbb{Z}^\times_N$$. Any set $$S=\{e_1,e_2,\ldots,e_n\}$$ can be commited via a commitment $$c=g^{\prod_{i\in[n]} \mathsf{H}(e_i)}\in\mathbb{Z}^\times_N$$ where $$\mathsf{H}:S\rightarrow\mathbb{P}$$ is a random oracle that maps the set $$S$$ to the set of primes $$\mathbb{P}$$. The memebership proof $$\pi_j=g^{\prod_{i\in[n]\setminus \{j\}} \mathsf{H}(e_i)}=c^{1/\mathsf{H}(e_j)}\in\mathbb{Z}^\times_N$$ allows anyone to verify $$e_j\in S$$ if and only if $$c=(\pi_j)^{\mathsf{H}(e_j)}$$.

My question is what if the commiter commits $$c=\pm1$$ dishonestly. Then for any odd prime $$\mathsf{H}(e_j)$$ the membership proof $$\pi_j=\pm 1$$ (respectively) will fool the verifier. Eliminating $$\pm 1$$ resolves the issue but does not it eliminate honestly computed commitments too? Or is there any reason to believe that the commitment $$c$$ can never be $$\pm 1$$ in the above setting?

It has been resolved using the discussion of the following question,
Is it possible to have square-free order(s) in $$\mathbb{Z}^\times_N$$?

In conclusion, the commitment $$c=\pm 1$$ can be eliminated safely with these three realistic assumptions,

i) $$N=(2p+1)(2q+1)$$ is a product of two safe primes $$p'=2p+1$$ and $$q'=2q+1$$.
ii) $$\mathsf{H}$$ maps each $$e_i$$ to odd primes except $$p$$ and $$q$$.
iii) The element $$g$$ is chosen to be $$1 and $$1.

Now, we prove that $$c\ne\pm 1$$ in the above setting.

Denote $$d$$ as the order of $$g$$, so $$d$$ divides the order of $$\mathbb{Z}^\times_N$$ that is $$4pq$$. Let, $$m=\prod_{i\in[n]} \mathsf{H}(e_i)$$ be the exponent. Since, $$\mathsf{H}(e_i)$$s are odd, $$m$$ is odd.

If $$g^m=1$$ then $$d$$ divides $$m$$. So, $$d$$ divides $$\mathsf{gcd}(m,4pq)=1$$ which implies $$d=1$$. In $$\mathbb{Z}^\times_N$$, the only element with order $$d=1$$ is the identity $$1$$ itself. But, $$g>1$$, so, $$c\ne 1$$.
Similarly, if $$g^m=-1$$ then $$g^{2m}=1$$, so $$d$$ divides $$2m$$. So, $$d$$ divides $$\mathsf{gcd}(2m,4pq)=2$$ which implies $$d=2$$. It means either $$g=-1 \pmod{p'}$$ or $$g=-1 \pmod{q'}$$ or both i.e., $$g=-1 \pmod{N}$$. Hence, $$g\ge p'-1$$ or $$g\ge q'-1$$. But, $$g< p'-1$$ and $$g< q'-1$$. Hence, $$c\ne-1$$.