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A nontrivial group has no proper nontrivial subgroups exactly when it is a cyclic group of prime group order. To prove necessity without presupposing finiteness, take $g\ne e$. Its cyclic subgroup is nontrivial and hence equals $G$. An infinite-order generator would give the proper nontrivial subgroup $⟨g^2⟩$, so its order $N$ is finite. If $N$ were composite, a proper divisor $d$ with $1<d<N$ would make $⟨g^d⟩$ another proper nontrivial subgroup. Thus $N$ is prime. Conversely Lagrange's theorem excludes intermediate subgroup orders in a prime-order group. The qualification “nontrivial” is necessary because every group with more than one element has the proper trivial subgroup $\{e\}$.

A cyclic group is generated by one element: $G=⟨g⟩=\{g^n:n\in \mathbb{Z}\}$. The hypothesis needs the qualification no proper nontrivial subgroup: every nontrivial group has the proper trivial subgroup $\{e\}$, so the literal wording admits no examples.

Under that intended hypothesis, choose $g\ne e$. The subgroup $⟨g⟩$ is nontrivial and must therefore be all of $G$, proving cyclicity. If $g$ had infinite order, then $⟨g^2⟩$ would be nontrivial and proper, since $g$ is not an even power of an infinite-order generator. Thus $g$ has finite order $N>1$ and $|G|=N$.

If $N$ were composite, choose a divisor $d$ with $1<d<N$. Then $⟨g^d⟩$ would have order $N/d$, strictly between one and $N$, again contradicting the hypothesis. ConsequentlyConversely, Lagrange's theorem shows that a group of prime order has only itself and its trivial subgroup. This proves the characterization of groups with no proper nontrivial subgroups, even without assuming finiteness at the outset.

For any cyclic subgroup $H=⟨s⟩$ of order $n\in \{1,2,4\}$, defineBoth preserve products because scalar phases and planar rotations add their angles. Each sends the generator to an element of exact order $n$, so both have trivial kernel and are faithful representations.

For $⟨U⟩\cong C_4$, take $s=U$: the generator images are $i$ and $R(\pi /2)$. For each order-two subgroup take its nonidentity element $s$; the images are $-1$ and $-I_2$. For the trivial subgroup the unique images are $1$ and $I_2$, which are faithful on a one-element group.

The one-dimensional representation of $C_4$ here must use complex scalars. There is no faithful one-dimensional real representation of $C_4$, since a nonzero real scalar of finite order is $1$ or $-1$, and neither has order four. The two-dimensional representations above are real.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,The left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ onto that coset, explaining the formula. If $|G|=p$ is prime, then $|H|$ is either $1$ or $p$, so $H=\{e\}$ or $H=G$. Thus the correct conclusion is no proper nontrivial subgroup. The trivial subgroup $\{e\}$ is proper for every group of prime order, so the wording in the question needs this qualification.