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The commutator subgroup equals the centre, soFix $\omega =e^{2\pi i/p}$. Every linear character isand these $p^2$ distinct characters exhaust the one-dimensional representations.

Take the abelian subgroupof order $p^2$. For each $c\in {\mathbb{F}}_p^{\times }$, define ${\theta }_c(x,0,z)={\omega }^{cz}$ and form the induced representationIt has degree $[G:A]=p$. Conjugation by representatives $(0,y,0)$ gives $p$ distinct characters of $A$ because $c\ne 0$. The stabilizer of ${\theta }_c$ is therefore exactly $A$, and the Mackey irreducibility criterion shows that ${\rho }_c$ is irreducible. The $p-1$ choices of $c$ have distinct central characters.

Because $K$ is a subcomplex, $\partial C_k(K)\subseteq C_{k-1}(K)$, sois well defined and squares to zero. The short exact sequence of chain complexesgives the long exact sequenceTaking $L=\partial {\Delta }^{n+1}$ and using the contractibility of ${\Delta }^{n+1}$ givesFor $0<k<n$, exactness makes $H_k(K)$ a quotient of a subgroup of $H_{k+1}(L,K)$, whose rank is at mostThe same stated inequality for $k=n$ follows from $H_n(K)\hookrightarrow H_n(L)\cong \mathbb{Z}$. Finally, an $n$-cycle in the boundary sphere has equal signed coefficients on all $n$-faces. A proper subcomplex omits an $n$-face, so that coefficient and hence every coefficient is zero. Therefore $H_n(K)=0$.

LetBoth have splitting field $\mathbb{Q}(\sqrt{2},\sqrt{3})$. On the roots of $f$, the Galois group has two orbits of size two and embeds asThe roots of $g$, namely $\pm \sqrt{2}\pm \sqrt{3}$, form one orbit and give the regular Klein-four subgroupThe subgroups are not conjugate because their orbit decompositions differ.

Dedekind's factorization/Frobenius theorem says that for a prime not dividing the discriminant, the degrees of the distinct irreducible factors modulo $p$ give the cycle lengths of an element of the Galois group.

The polynomial is irreducible over $\mathbb{Q}$ by the rational-root test followed by a comparison of possible monic quadratic factors. Its discriminant is $547^2$, a square, so its transitive Galois group lies in $A_4$. Modulo five,and the cubic has no root in ${\mathbb{F}}_5$, giving a three-cycle. A transitive subgroup of $A_4$ containing a three-cycle cannot be the Klein four group; hence

The centre isEvery subgroup $H\le Z(G)$ is normal, since $gh{g}^{-1}=h$ for all $g\in G$ and $h\in H$.

Write $D_{2n}=⟨r,s:r^n=s^2=1,srs=r^{-1}⟩$. A central rotation must satisfy $r^k=r^{-k}$, and no reflection is central for $n\ge 3$. Hence

False. The proper subgroup $2\mathbb{Z}\times 2\mathbb{Z}$ of ${\mathbb{Z}}^2$ is isomorphic to ${\mathbb{Z}}^2$ and is not cyclic.

If $N◃S_n$, then $N\cap A_n◃A_n$, hence the intersection is either $1$ or $A_n$. In the latter case $N$ is $A_n$ or $S_n$. In the former, $N$ injects into $S_n/A_n\cong C_2$, so $|N|\le 2$. A normal subgroup of order two would be central, but $Z(S_n)=1$ for $n\ge 3$. Therefore the normal subgroups are

With $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$, its normal subgroups areThe last three proper nontrivial examples have index two; the four individual reflection subgroups are not normal.

False. The quaternion group $Q_8$ is nonabelian, but each of its subgroups is normal: its nontrivial proper subgroups are its centre $\{\pm 1\}$ and the three cyclic subgroups of order four, all of index two.

Invertible matrices are closed under multiplication, contain $I$, have associative multiplication, and have inverses by the adjugate formula over the field ${\mathbb{F}}_p$. The first column can be any nonzero vector and the second any vector outside its span, so

For $p=2$, the matrix $\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ has order three. There is no element of order six: $G{L}_2({\mathbb{F}}_2)\cong S_3$, whose element orders are $1,2,3$.

For $p>2$, $S{L}_2({\mathbb{F}}_p)=\ker \det$ is a proper normal subgroup. For $p=2$, the order-three subgroup in the copy of $S_3$ is proper and normal.

In $G{L}_2({\mathbb{F}}_{11})$, takewhere $A\le {\mathbb{F}}_{11}^{\times }$ is the subgroup of order five. This semidirect product has order $55$ and is nonabelian because a nontrivial diagonal element does not commute with translations.

Let $k=|G/H|$. The coset action gives a nontrivial homomorphism $G\to S_k$ whose kernel is normal. Simplicity makes it injective. The sign map is trivial on the nonabelian simple image, so $G$ embeds in $A_k$. For $k\le 4$, $A_k$ is solvable, as are its subgroups, whereas a nonabelian simple group is not. Hence $k\ge 5$.

Sylow gives $n_7\mid 24$ and $n_7\equiv 1(\mathrm{mod}7)$. Simplicity excludes $n_7=1$, so $n_7=8$. Distinct order-seven subgroups intersect trivially, hence there areelements of order seven.

If a finite group is not simple, choose a maximal proper normal subgroup and repeat inside it. Orders strictly decrease, so the process terminates; maximality makes every quotient simple.

For $S_4$ one composition series isThe successive quotients have orders $2,3,2,2$, hence are simple. As subgroups of $S_4$, the groups $S_4,A_4,V_4$, and $1$ are normal. The selected order-two subgroup is not normal, since conjugation permutes the three double transpositions.