Question

If $\mathit{G}$ is a group of order$\mathit{n}$ and$\mathit{G}$ has${\mathbf{2}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$ subgroups, prove that $\mathit{G}\mathbf{=}\mathbf{⟨}\mathbf{e}\mathbf{⟩}$or$\mathit{G}\mathbf{\cong }{\mathit{\mathbb{Z}}}_{\mathbf{2}}$ .

Short answer

It is proved that, $\mathrm{G}=\left\{\mathrm{e}\right\}\mathrm{or}\mathrm{G}\cong {\mathrm{\mathbb{Z}}}_2$.

Step-by-step solution

Determine G=⟨e⟩ or G≅ℤ2

Consider the given function,$G$ is the group of order$n$and$G$has$2^{n-1}$subgroups.

Let’s assume that,$G\ne ⟨e⟩$. Then $G/\left\{e\right\}$has $n-1$elements and thus it is exactly $2^{n-1}$subset.

The hypothesis is that all subsets together with the identity$e$ are the subgroups of $G$.

That means any subset$S$ of$G/\left\{e\right\},S\cup \left\{e\right\}$ is the subgroup of $G$.

Determine further simplification of G=⟨e⟩or G≅ℤ2

Assume that, $\left|G\right|>2$. Consider role="math" localid="1654352720274" $\{a,b\}\subset G/\left\{e\right\}$, here role="math" localid="1654352739221" $a\ne b\ne e$then the hypothesis, both $\{a,e\}$and $\{e,a,b\}$are the subgroups of $G$. Then it follows that,role="math" localid="1654352843963" $a^2=e$and $a^3=e$.

As $\{a,e\}$is a group and the order of grouprole="math" localid="1654352973401" $\{e,a,b\}$ is 3. Then$b=a^2=e$ thus$b\ne e$ ,

which implies $G=⟨e⟩$.

Which is a contradiction to $G\ne ⟨e⟩$.

Hence, proved.