Chapter 3 Abstract Algebra (1)


** Discrete Mathematics **


Prove that multiplication makes $\mathbb{Q}(\sqrt{2}) \backslash\{0\}$ into a group.

Show that if $x$ and $y$ are elements of a group, prove that $(x y)^{-1}=y^{-1} x^{-1}$.

Let G be a cyclic group. Show that every subgroup of G is cyclic.

If $G$ is a cyclic group and $N$ is a subgroup, prove that $G / N$ is cyclic.

Find (up to isomorphism) all abelian groups of order $72 .$

If $G$ is an Abelian group of order $p_{1} p_{2} \ldots p_{k}$ where the $p_{i}$ are distinct primes, prove that $G$ is cyclic.

(1st Isomorphism Theorem). Let $\phi: G \rightarrow H$ be a homomorphism. Then $\operatorname{Im} \phi \leq H, \operatorname{Ker} \phi \leq G$ and
$$ G / \operatorname{Ker} \phi \cong \operatorname{Im} \phi $$

(2nd Isomorphism Theorem). Let $H \leq G$ and $N \leq G .$ Then $H N \leq G$, $H \cap N \leq H$ and
$$ H /(H \cap N) \cong H N / N $$

(3rd Isomorphism Theorem). Suppose that $H, N \leq G$ and $N \leq H .$ Then $H / N \leq G / N$ and
$$ (G / N) /(H / N) \cong G / H . $$

Let $G$ be a non-Abelian group of order $21, P$ a Sylow 7 -subgroup and $Q$ a Sylow 3 subgroup of $G$ pts)
(a) Show that $P$ and $Q$ are cyclic.
(b) Show that $P$ is a normal subgroup of $G$.
(c) Show that $Q$ is not a normal subgroup of $G$.

Let $p$ be a prime and $P$ a group of order $p^{2}$. Show that
$P$ is Abelian.

Explain that every Abelian group of order 675 has at least 8 elements of order 15.

Prove that a group of order 150 is not simple.

Let $I$ and $J$ be ideals of a ring $R$. Prove that the set $I+J$ of elements of the form $x+y$, with $x$ in $I$ and $y$ in $J$, is an ideal.

Let $I$ and $J$ be ideals of a ring $R$. Prove that the intersection $I \cap J$ is an ideal.

Are the rings $\mathbb{Z}[x] /\left(x^{2}+7\right)$ and $\mathbb{Z}[x] /\left(2 x^{2}+7\right)$ isomorphic?

Show that the $ PIDs$ must be $ UFDs$.


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

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

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 $\left\{a^{i} \mid i \in \mathbb{Z}\right\}=\langle a\rangle=G .$ Now,
$$ G / N=\{b N \mid b \in G\}=\left\{a^{i} N \mid i \in \mathbb{Z}\right\}=\left\{(a N)^{i} \mid i \in \mathbb{Z}\right\}=\langle a N\rangle $$

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

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
$$ \mathbb{Z}_{8} \oplus \mathbb{Z}_{9}, \quad \mathbb{Z}_{2} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_{9}, \quad \mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{9}, $$
$$ \mathbb{Z}_{8} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3}, \quad \mathbb{Z}_{2} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3}, \quad \mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} $$

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 $a=a_{1} \ldots a_{k} .$ Set $\hat{p}_{i}=\left(p_{1} \ldots p_{k}\right) / p_{i}$ Certainly the order of $a$ divides $p_{1} p_{2} \ldots p_{k}$ by Lagrange's theorem.

Note that $a^{\bar{p}_{i}}=e a^{\bar{p}_{i}} e .$ Note that $a^{\dot{p} i} \neq e$ 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 $p_{1} p_{2} \ldots p_{k} .$ This implies that $G$ is cyclic as desired.

Solution of 8:
Proof We apply the 1st Isomorphism Theorem. Consider the homomorphism
$$ \phi: G \rightarrow G / N, a \mapsto a N $$
Let $\psi$ be the restriction of $\phi$ on $H .$ This gives us a homomorphism $\psi: H \rightarrow G / N .$ By the 1st Isomorphism Theorem we have that $\operatorname{Im} \psi=\{h N: h \in H\}$ 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
$$ U=\bigcup_{h \in H} h N=H N $$
Thus $\operatorname{Im} \psi=H N / N$. It remains to identify the kernel. The identity of $G / N$ is the coset $e N=N$. Then for $h \in H,$ we have
$$ \begin{aligned} \psi(h)=N & \Leftrightarrow h N=N \\ & \Leftrightarrow h \in N \end{aligned} $$
As $h \in H$ this shows that the kernel of $\psi$ is $H \cap N$. Thus by the 1 st Isomorphism Theorem, $H \cap N \triangleleft H$ and
$$ H / H \cap N=H / \operatorname{Ker} \psi \simeq \operatorname{Im} \psi=H N / N $$
This finishes the proof. $\square$

Solution of 9:

Proof Again we apply the 1st Isomorphism Theorem. This time on the map
$$ \phi: G / N \rightarrow G / H a N \mapsto a H . $$
Let us first see that this is well defined. If $a N=b N$ then $a^{-1} b \in N \subseteq H$ and thus $a H=b H .$ It is also a homomorphism as
$$ \phi(a N \cdot b N)=\phi(a b N)=a b H=a H \cdot b H=\phi(a N) \cdot \phi(b N) $$
We clearly have that $\operatorname{Im} \phi=G / H$ and it remains to identify the kernel. The identity in $G / H$ is the coset $e H=H$ and then
$$\begin{aligned} \phi(a N)=H & \Leftrightarrow a H=H \\ & \Leftrightarrow a \in H . \end{aligned} $$
The kernel thus consists of the cosets $a N$ of $G / N$ where $a \in H .$ That is the kernel is $H / N .$ The 1st Isomorphism Theorem now gives us that $H / N \leq G / N$ (that we had proved already in the proof of the correspondence theorem anyway) and that
$$ (G / N) /(H / N)=(G / N) / \operatorname{Ker} \phi \cong \operatorname{Im} \phi=G / H $$
This finishes the proof. $\square$

Solution of 10:

(a) Every groups of prime order are cyclic. In fact,
let $|G|=p$, where $p$ is a prime. Let $e \neq x \in G$. Then $H=\langle x\rangle$ 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, $|G: N(P)|=\left|\operatorname{Syl}_{7}(G)\right| \equiv 1 \quad(\bmod 7)$. 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 $P Q$ is a subgroup of $G$ of order divisible by 7 and 3 , we have $G=P Q$ 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 $R=\langle x\rangle$. Then both $Q$ and $R$ are normal and $Q \cap R=\{e\}$ as $x \notin Q$. Threfore $Q R=Q \times R \leq P$. 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 $\Sigma$ be the set of all Sylow 5 -groups in $G$.
Consider the permutation representation $\rho$ of $G$ on $\Sigma$ which sends $P$ to $g P g^{-1}$ for each $P \in \Sigma$ and $g \in G$. By Sylow Theorem, $G$ acts transitively on $\Sigma$. The stabilizer subgroup of $H$ in $G$ is $\left\{g \in G \mid g H g^{-1}=H\right\}=N$. Thus $|\Sigma|=[G: N] \equiv 1(\bmod 5)$
Since $H \triangleleft N \leqslant G$, we have $[G: N]=\frac{[G: H]}{[N: H]}=\frac{6}{[N: H]} .$
$[G: N]=1$ or $[G: N]=6$. In the case $[G: N]=1, H$ is a normal subgroup of $G$, so $G$ is not simple. Suppose $|\Sigma|=[G: N]=6$. We have $[N: H]=1$ and the kernel of $\rho$ is $\cap_{g \in G} g H g^{-1}$. If $\operatorname{ker}(\rho)$ is is non-trivial, then $G$ is not simple. If $\operatorname{ker}(\rho)$ is trivial, then $\rho$ induces an embedding $\rho: G \hookrightarrow \mathrm{Sym}_{\Sigma} \cong S_{6} .$ 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 $(x+y),\left(x^{\prime}+y^{\prime}\right) \in I+J$ and $s \in R$, where $x, x^{\prime} \in I, y, y^{\prime} \in J .$ Then, $s(x+y)+\left(x^{\prime}+y^{\prime}\right)=\left(s x+x^{\prime}\right)+\left(s y+y^{\prime}\right) \in I+J$, hence $I+J$ is an ideal.

Solution of :

We claim they are not. Let $R=\mathbb{Z}[x] /\left(x^{2}+7\right)$ and $S=\mathbb{Z}[x] /\left(2 x^{2}+7\right) .$ Asume $\alpha: R \rightarrow S$ 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, $R /(2) \approx F_{2}[x] /\left(x^{2}+1\right) \neq 0$, whereas $S /(2) \approx F_{2}[x] /(1)=0$







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