Chapter 3 Abstract Algebra (1)

Solution of 2:
The element $(xy{)}^{-1}$ is the element s.t. $(xy{)}^{-1}(xy)=1=(xy)(xy{)}^{-1}.$ We show that the element $y^{-1}{x}^{-1}$ satisfies this property.
$$\begin{array}{rl}\left(y^{-1}{x}^{-1}\right)(xy)& =y^{-1}(x^{-1}(xy))\\ & =y^{-1}\left(\left(x^{-1}x\right)y\right)\\ & =y^{-1}y\\ & =1\end{array}$$
so the first part of that equation is satisfied. Now for the second part.
$$\begin{array}{rl}(xy)\left(y^{-1}{x}^{-1}\right)& =((xy)y^{-1})x^{-1}\\ & =\left(x\left(y{y}^{-1}\right)\right)x^{-1}\\ & =x^{-1}x\\ & =1\end{array}$$

then we have $(xy{)}^{-1}=y^{-1}{x}^{-1}$.

Solution of 4:

First note that $N$ is normal since $G$ being cyclic implies that $G$ is Abelian so the question makes sense. Suppose that $\{a^i\mid i\in \mathbb{Z}\}=⟨a⟩=G.$ Now,
$$G/N=\{bN\mid b\in G\}=\{a^{i}N\mid i\in \mathbb{Z}\}=\{(aN{)}^i\mid i\in \mathbb{Z}\}=⟨aN⟩$$

This proves that $G/N$ is cyclic generated by $aN.$

Solution of 5:
We have $72=2^3\cdot 3^2$. The possible partitions of $2^3$ are (8),(2,4),(2,2,2) whereas the possible partions for $3^2$ are $\left(3^2\right),(3,3)$. We then have that the abelian groups of order 72 are
$${\mathbb{Z}}_8\oplus {\mathbb{Z}}_9,{\mathbb{Z}}_2\oplus {\mathbb{Z}}_4\oplus {\mathbb{Z}}_9,{\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\oplus {\mathbb{Z}}_9,$$
$${\mathbb{Z}}_8\oplus {\mathbb{Z}}_3\oplus {\mathbb{Z}}_3,{\mathbb{Z}}_2\oplus {\mathbb{Z}}_4\oplus {\mathbb{Z}}_3\oplus {\mathbb{Z}}_3,{\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\oplus {\mathbb{Z}}_3\oplus {\mathbb{Z}}_3$$

Solution of 6:

If $k=1$ we already know the answer, so we can suppose $k>1$. We know that $G$ contains an element $a_i\in G$ of order $p_i$. Consider the element $a=a_1\dots a_k.$ Set ${\hat{p}}_i=(p_1\dots p_k)/p_i$ Certainly the order of $a$ divides $p_1{p}_2\dots p_k$ by Lagrange's theorem.

Note that $a^{{\overline{p}}_i}=e{a}^{{\overline{p}}_i}e.$ Note that $a^{\dot{p}i}\ne e$ since $p_i$ does not divide ${\hat{p}}_i$ (and $a_i$ has order $p_i$ ). But this implies that the order of $a$ is bigger than ${\hat{p}}_i$ for each $i$. Since the order of $a$ divides the product of all the $p_i$, this implies that the order of $a$ must equal $p_1{p}_2\dots p_k.$ This implies that $G$ is cyclic as desired.

Solution of 8:
Proof We apply the 1st Isomorphism Theorem. Consider the homomorphism
$$\varphi :G\to G/N,a\mapsto aN$$
Let $\psi$ be the restriction of $\varphi$ on $H.$ This gives us a homomorphism $\psi :H\to G/N.$ By the 1st Isomorphism Theorem we have that $\mathrm{Im}\psi =\{hN:h\in H\}$ is a subgroup of $G/N$. By the correspondence theorem we have that this subgroup is of the form $U/N,$ where $U$ is a subgroup of $G$ that is given by
$$U=\bigcup _{h\in H}hN=HN$$
Thus $\mathrm{Im}\psi =HN/N$. It remains to identify the kernel. The identity of $G/N$ is the coset $eN=N$. Then for $h\in H,$ we have
$$\begin{array}{rl}\psi (h)=N& \iff hN=N\\ & \iff h\in N\end{array}$$
As $h\in H$ this shows that the kernel of $\psi$ is $H\cap N$. Thus by the 1 st Isomorphism Theorem, $H\cap N◃H$ and
$$H/H\cap N=H/\mathrm{Ker}\psi \simeq \mathrm{Im}\psi =HN/N$$
This finishes the proof. $◻$

Solution of 9:

Proof Again we apply the 1st Isomorphism Theorem. This time on the map
$$\varphi :G/N\to G/HaN\mapsto aH.$$
Let us first see that this is well defined. If $aN=bN$ then $a^{-1}b\in N\subseteq H$ and thus $aH=bH.$ It is also a homomorphism as
$$\varphi (aN\cdot bN)=\varphi (abN)=abH=aH\cdot bH=\varphi (aN)\cdot \varphi (bN)$$
We clearly have that $\mathrm{Im}\varphi =G/H$ and it remains to identify the kernel. The identity in $G/H$ is the coset $eH=H$ and then
$$\begin{array}{rl}\varphi (aN)=H& \iff aH=H\\ & \iff a\in H.\end{array}$$
The kernel thus consists of the cosets $aN$ of $G/N$ where $a\in H.$ That is the kernel is $H/N.$ The 1st Isomorphism Theorem now gives us that $H/N\le G/N$ (that we had proved already in the proof of the correspondence theorem anyway) and that
$$(G/N)/(H/N)=(G/N)/\mathrm{Ker}\varphi \cong \mathrm{Im}\varphi =G/H$$
This finishes the proof. $◻$

Solution of 10:

(a) Every groups of prime order are cyclic. In fact,
let $|G|=p$, where $p$ is a prime. Let $e\ne x\in G$. Then $H=⟨x⟩$ is a subgroup of $G$ of order at least 2 . By Lagrange's Theorem, $|H|=|G|$ and $H=G$. Hence $G$ is cyclic.
Since $|P|=7$ and $|Q|=3$, both $P$ and $Q$ are cyclic as 7 and 3 are prime.

(b) By Sylow's Theorem, $|G:N(P)|=|{\mathrm{Syl}}_7(G)|\equiv 1(mod7)$. Since $|G:N(P)|$ is
a divisor of $|G|$ by Lagrange's Theorem, we have $|G:N(P)|=1$ and $G=N(P)$. By the definition of $N(P),G=N(P)$ means $P$ is a normal subgroup.

(c) We want to show it by way of contradiction, assume that $Q$ is a normal subgroup. Then $|P\cap Q|$ is a divisor of $|P|$ and $|Q|$ and $P\cap Q=\{e\}$. Now $PQ$ is a subgroup of $G$ of order divisible by 7 and 3 , we have $G=PQ$ and $G=P\times Q.$ By (a), both $P$ and $Q$ are cyclic, so Abelian, $G$ is Abelian. This contradicts our assumption. Thus $Q$ is not normal in $G$.

Solution of 11:

We may assume that $P$ is not cyclic. Hence every nonidentity element of $P$ generates a cyclic subgroup of order $p$ .

Let $Q$ be a subgoup of order $p$, and $x\notin Q$. Then $x$ is of order $p$ again, as $P$ is not cyclic. Let $R=⟨x⟩$. Then both $Q$ and $R$ are normal and $Q\cap R=\{e\}$ as $x\notin Q$. Threfore $QR=Q\times R\le P$. By comparing their orders, we have $P=Q\times R$ and $P$ is Abelian.

Solution of 12 :

Since G is Abelian, for each divisor m of its order, there is a subgroup of order m. Since the only Abelian group of order 15 is cyclic, there is an element of order 15. Since φ(15) = φ(3)φ(5) = 2 · 4 = 8, there are 8 elements of order 8 in a cyclic group of order 15. Therefore, there are at least 8 elements of order 15.

Solution of 13 :

Proof. Note that $150=2\times 3\times 5^2$. We have that by Sylow Theorem, Sylow 5-groups exist, mutually conjugate and the number of Sylow 5groups is congruent to 1 modulo $5.$ Let $H$ be a Sylow 5 -group and $N=N_G(H).$ Let $\mathrm{\Sigma }$ be the set of all Sylow 5 -groups in $G$.
Consider the permutation representation $\rho$ of $G$ on $\mathrm{\Sigma }$ which sends $P$ to $gP{g}^{-1}$ for each $P\in \mathrm{\Sigma }$ and $g\in G$. By Sylow Theorem, $G$ acts transitively on $\mathrm{\Sigma }$. The stabilizer subgroup of $H$ in $G$ is $\{g\in G\mid gH{g}^{-1}=H\}=N$. Thus $|\mathrm{\Sigma }|=[G:N]\equiv 1(mod5)$
Since $H◃N⩽G$, we have $[G:N]=\frac{[G:H]}{[N:H]}=\frac{6}{[N:H]}.$
Therefore
$[G:N]=1$ or $[G:N]=6$. In the case $[G:N]=1,H$ is a normal subgroup of $G$, so $G$ is not simple. Suppose $|\mathrm{\Sigma }|=[G:N]=6$. We have $[N:H]=1$ and the kernel of $\rho$ is ${\cap }_{g\in G}gH{g}^{-1}$. If $\ker (\rho )$ is is non-trivial, then $G$ is not simple. If $\ker (\rho )$ is trivial, then $\rho$ induces an embedding $\rho :G\hookrightarrow {\mathrm{Sym}}_{\mathrm{\Sigma }}\cong S_6.$ However $\left|S_6\right|=6!=5\times 3^2\times 2^4$
and 25 divides $|G|$, which contradicts to Lagrange Theorem on orders of subgroups.

Solution of 14:

Proof. Let $(x+y),(x^{\prime }+y^{\prime })\in I+J$ and $s\in R$, where $x,x^{\prime }\in I,y,y^{\prime }\in J.$ Then, $s(x+y)+(x^{\prime }+y^{\prime })=(sx+x^{\prime })+(sy+y^{\prime })\in I+J$, hence $I+J$ is an ideal.

Solution of :

We claim they are not. Let $R=\mathbb{Z}[x]/(x^2+7)$ and $S=\mathbb{Z}[x]/(2x^2+7).$ Asume $\alpha :R\to S$ is an isomorphism. $\alpha (1)=1$, hence $\alpha (2)=2$, and so if $\alpha$ is an isomorphism, then $R/(2)\approx S/(2)$. We claim this is a contradiction. For, $R/(2)\approx F_2[x]/(x^2+1)\ne 0$, whereas $S/(2)\approx F_2[x]/(1)=0$