Groups

A finite group is the union of its proper subgroups iff it is not cyclic: if it is cyclic a generator lies in no proper subgroup; if not cyclic every element generates a proper subgroup.

(1) and (2) form the subgroup structure theorem for $\mathbb{Z}/n\mathbb{Z}.(3)$ fails: a single rational a/b cannot generate $\mathbb{Q}$, since a/(2b) is never an integer multiple of a/b. (4) Lagrange: any non-identity element generates a subgroup of order dividing p and exceeding 1, hence the whole group.

Generators of $\mathbb{Z}/n$ are the residues coprime to n, counted by Euler${}^{\prime }s\phi :\phi (12)=12(1-1/2)(1-1/3)=4$, namely 1, 5, 7, 11.

A p-group of order $p^n$ has a normal subgroup of every order pᵏ for 0 ≤ k ≤ n, giving at least n + 1 = 6.

For g ∉ H, the left coset gH and the right coset Hg both equal G \ H — with only two cosets there is nowhere else to go — so gH = Hg for every g. No abelian or finiteness hypothesis is needed: $A_n$ in $S_n$ is the standard instance.

(b) fails for $Q_8.(c)$ fails for ⟨(12)(34)⟩ ⊴ $V_4$ ⊴ $A_4$.

Order 6 needs cycle type (3,2): 5!/(3·2) = 20.

$A_3\cong {\mathbb{Z}}_3$ is simple but $A_4$ is not $(V_4$ is normal); simplicity holds for n ≥ 5.

(1) The order of a product of disjoint cycles is the lcm of the lengths: lcm(3, 2) = 6. (2) A 3-cycle is even, a transposition odd, so the product is odd. (3) Conjugacy in $S_n$ is exactly equality of cycle type, and both have type 3 + 2. (4) fails: $A_5$ is the even permutations.

A finite p-group has a normal subgroup of every order pᵏ dividing |G|, so |G| $=p^6$ gives at least seven normal subgroups (orders $1,p,\dots ,p^6)$ — 'exactly six' is impossible. (1), (2) follow from the same fact; (3) an infinite abelian p-group (e.g. the Prüfer group) has infinite centre.

(1) In the class equation |G| = |Z(G)| $+\sum [G:C(x_i)]$, every index on the right is a positive power of p, so p divides |Z(G)|. (2) G/Z(G) has order 1 or p, hence is cyclic, and G/Z cyclic forces G abelian. (3) fails: $\mathbb{Z}/p\times \mathbb{Z}/p.(4)$ The non-trivial abelian group Z(G) contains an element of order p; the subgroup it generates is central, hence normal in G.

$2023=7\cdot 17^2.n_7\equiv 1(mod7)$ and $n_7$ | $289\Rightarrow n_7=1;n_{17}$ | 7 and $\equiv 1(mod17)\Rightarrow n_{17}=1$. So $G\cong {\mathbb{Z}}_7\times P$ with P of order $17^2$ abelian: G is abelian and not simple. P may be ${\mathbb{Z}}_{17}\times {\mathbb{Z}}_{17}$, so G need not be cyclic.

$n_5\equiv 1(mod5)$ and $n_5$ | $4\Rightarrow n_5=1$.

Here $n_3\in$ {1, 4} and $n_2\in$ {1, 3}. (1) If $n_3=4$ then the four Sylow 3-subgroups contribute 4 × 2 = 8 elements of order 3, leaving only 4 elements, which must form the unique — hence normal — Sylow 2-subgroup. So some Sylow subgroup is always normal. (2) Fails for the dihedral group of order 12, where $n_2=3.(3)$ True: every group of order 12 has a normal Sylow subgroup by (1), and the quotient and subgroup are of order ≤ 4, so G is solvable. (4) Fails for $A_4$, the standard example of a group with no subgroup of index 2.

$n_5$ divides 3 and is ≡ 1 (mod 5), so $n_5=1;n_3$ divides 5 and is ≡ 1 (mod 3), so $n_3=1$. Both Sylow subgroups are normal, intersect trivially, and their product is the whole group: $G\cong \mathbb{Z}/3\times \mathbb{Z}/5\cong \mathbb{Z}/15$. Having normal subgroups it is not simple.

$72=2^3\cdot 3^2:p(3)\cdot p(2)=3\cdot 2=6$.

There are $p^2-1$ elements of order p, each subgroup contains p − 1 of them, giving $(p^2-1)/(p-1)=p+1$.

(1) Partitions of the exponent 3 for the 2-part: 8, 4×2, 2×2×2. (2) 12 = 4·3 or 2·2·3. (3) By CRT, $\mathbb{Z}/6\times \mathbb{Z}/4\cong \mathbb{Z}/2\times \mathbb{Z}/3\times \mathbb{Z}/4\cong (\mathbb{Z}/4\times \mathbb{Z}/3)\times \mathbb{Z}/2\cong \mathbb{Z}/12\times \mathbb{Z}/2.(4)$ fails: the left side has an element of order 8, while the right side has exponent 4.