• #todo/school : Do Problems 2.1: 1b, 2d, 4, 8, 9, 10, 15 📅 2025-01-26 ✅ 2025-01-23

1b

Proof

To show $C=\{z\in \mathbb{C}||z|=1\}$ is a subgroup under multiplication, we must show closed under the operation, and closure under inverses.

1. Let $a,b\in C$. Then $a=a_1+i{a}_2$ and $b=b_1+i{b}_2$ where $a_1^2+a_2^2=1$ and $b_1^2+b_2^2=1$. Then notice:

To show that $ab\in C$ notice that:

Thus $ab\in C$.

2. Let $a\in C$ be as described in (1). I claim that $\bar{a}=a_1-i{a}_2$ is the inverse in this case. Notice that this is normally the case for $\mathbb{C}$, just without the magnitude scaling factor since $|z|=1$:

which is the identity. To show closure, notice that:

Thus $\bar{a}\in C$.

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2d

Proof

Denote the subset $Z=\{2n+1|n\in \mathbb{Z}\}\cup \{0\}$ with the usual operation of addition. Namely, we have the inverse (additive inverses of odd integers are odd) so we are missing closure. Namely, the counter example $1,3\in Z$ while $1+3=4\notin Z$ shows that $+{|}_Z$ isn't closed, so $(Z,+{|}_Z)$ is not a valid subgroup.

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4

Proof

Let the binary group operation be addition $+$. Let $G=\mathbb{R}$. Let $H={\mathbb{R}}^{+}$ so then $H\subseteq \mathbb{R}$. Notice that $H$ is infinite and $(G,+)$ is a group, but $(H,+)$ is not a subgroup, as $H$ is missing the identity element, namely $0\notin H$. However, the operation $+{|}_H$ is closed as any positive real plus a positive real must be positive itself.

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8

Proof

$(\Rightarrow )$: Suppose $H\cup K$ is a subgroup. If $H\subseteq K$ already then we are done so suppose instead $H⊈K$ instead. Let $k\in K$ be arbitrary, so we can show that $k\in H$ and thus $K\subseteq H$.

Since $k\in K$ then $k\in H\cup K$. Now notice that since $H⊈K$ then $\mathrm{\exists }{k}^{\prime }\in H$ where $k^{\prime }\notin K$. Now clearly $k^{\prime }\in H\cup K$, so since $H\cup K$ is a subgroup then $k^{\prime }k\in H\cup K$.

Now if $k^{\prime }k\in H$ then we are done since $k^{\prime }\in H$ so then $k\in H$ (if instead $k\notin H$ that would contradict $H$ being a subgroup). So consider the case where $k^{\prime }k\notin H$. Then $k^{\prime }k\in K$. But notice if this was the case then since $k\in K$ and $K$ is a subgroup then $k^{-1}\in K$. Since $K$ is a subgroup then $k^{\prime }k(k^{-1})=k^{\prime }\in K$, which is a contradicion. Thus then $k^{\prime }k\in H$ is the only case, and thus $k\in H$ is always going to happen.

Thus $K\subseteq H$ as desired.

$(\Leftarrow )$: Suppose either $H\subseteq K$ or $K\subseteq H$ is true. Suppose, WLOG, that $H\subseteq K$ (the other version is proved reflexively). To show $H\cup K$ is a subgroup, we just want to show that the operation on the set is closed and each element has its inverse in the set.

For the former, let $h,h^{\prime }\in H\cup K$. We have 4 cases:

• If $h,h^{\prime }\in H$ then by $H$ being a subgroup then $h{h}^{\prime }\in H\cup K$
• If $h\in H$ and $h^{\prime }\notin H$ then $h^{\prime }\in K$. Notice then $h\in K$ since $H\subseteq K$. Thus $h{h}^{\prime }\in K$ since $K$ is a subgroup. Thus $h{h}^{\prime }\in H\cup K$.
• If $h\notin H$ and $h^{\prime }\in H$ then the same argument is repeated (swap $h$ with $h^{\prime }$).
• If $h,h^{\prime }\notin H$ then $h,h^{\prime }\in K$. Thus $h{h}^{\prime }\in K$ by $K$ being a subgroup, so then $h{h}^{\prime }\in H\cup K$.
  No matter what $h{h}^{\prime }\in H\cup K$ as desired.

For the latter, let $h\in H\cup K$. We have two cases:

• If $h\in H$ then $h^{-1}\in H$ by $H$ being a subgroup. Thus $h^{-1}\in H\cup K$.
• If $h\notin H$ then $h\in K$. So $h^{-1}\in K$ by $K$ being a subgroup. Thus $h^{-1}\in H\cup K$.
  Thus no matter what we have $h^{-1}\in H\cup K$ as desired.

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9

Proof

First, for showing closure under matrix multiplication, let $A,B\in S{L}_n(\mathbb{F})$. Then $\text{det}(A)=\text{det}(B)=1$. Consider $AB$. Notice that:

Thus since $A,B$ are $n\times n$ and result in an $n\times n$ matrix, then $AB\in S{L}_n(\mathbb{F})$, showing closure.

For showing the inverse exists, $A^{-1}$ always exists for $A\in S{L}_n(\mathbb{F})$ since $det(A)=1\ne 0$. Furthermore, $A^{-1}$ must be $n\times n$ as inverse square matrices are the same dimension. Thus $A^{-1}\in S{L}_n(\mathbb{F})$.

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10

Proof

a. Let $H,K\le G$. Consider $H\cap K$.

First, for closure of the operation on $H\cap K$, let $a,b\in H\cap K$. Then $a,b\in H;a,b\in K$. Now because $H,K$ are subgroups then $ab\in H$ and $ab\in K$. Thus $ab\in H\cap K$, showing closure.

Second, for the existence of the inverse in $H\cap K$, let $a\in H\cap K$. Then $a\in H$ and $a\in K$. Since both $H,K$ are subgroups, then $a^{-1}\in H$ and $a^{-1}\in K$. Thus $a^{-1}\in H\cap K$ as desired.

b. Let $\mathcal{A}=\{G_{\lambda }|\lambda \in \mathrm{\Lambda }\}$ be a collection of groups (so $G_{\lambda }$ is a group for each $\lambda \in \mathrm{\Lambda }$).

Consider the set $G^{\prime }=\bigcap _{\lambda \in \mathrm{\Lambda }}{G}_{\lambda }$. To show $G^{\prime }$ is a subgroup, still $G^{\prime }\subseteq G$ so we just show the same two properties.

First, for closure of the operation on $G^{\prime }$, let $a,b\in G^{\prime }$. Then $a,b\in G_{\lambda }$ for all $\lambda \in \mathrm{\Lambda }$. Since $G_{\lambda }$ is a group, then $ab\in G_{\lambda }$. Notice then that since $\lambda \in \mathrm{\Lambda }$ was arbitrary, then $ab\in \bigcap _{\lambda \in \mathrm{\Lambda }}{G}_{\lambda }$, so then $ab\in G^{\prime }$. Thus $G^{\prime }$ is closed under the operation.

Second, let $a\in G^{\prime }$. Then $a\in G_{\lambda }$ for all $\lambda \in \mathrm{\Lambda }$. Now since $G_{\lambda }$ is a group then $a^{-1}\in G_{\lambda }$. Now because $\lambda$ is arbitrary, then $a^{-1}\in \bigcap _{\lambda \in \mathrm{\Lambda }}{G}_{\lambda }$, so then $a^{-1}\in G^{\prime }$, showing closure of the inverse.

Thus $G^{\prime }$ is a subgroup of $G$.

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15

Proof

Denote $H=\bigcup _{i=1}^{\mathrm{\infty }}{H}_i$. To show it's a subgroup of $G$, we follow by the conditions for being a subgroup.

First, let $h,k\in H$ be arbitrary. Then $\mathrm{\exists }{H}_{n_h},H_{n_k}$ where $n_h,n_k\in \mathbb{N}$ and $h\in H_{n_h}$ and $k\in H_{n_k}$. Now if $n_h=n_k=n$ then $h,k\in H_n$, so then by $H_n$ being a subgroup of $G$ then $hk\in H_n$ and thus $hk\in H$ (this is the trivial case). If $n_h\ne n_k$ then one of these is larger. WLOG say $n_h>n_k$. Then notice that $H_{n_k}\le H_{n_h}$ by our ascending chain:

As a result, then since $k\in H_{n_k}\le H_{n_h}$ then by extension then $k\in H_{n_h}$. Additionally $h\in H_{n_h}$ so then by closure of our binary operation for $H_{n_h}$ by being a subgroup, then $hk\in H_{n_h}$. Thus then $hk\in H$ by definition.

No matter what, we always get $hk\in H$, thus showing closure of the operation on $H$.

Second, let $h\in H$ be arbitrary. Then $\mathrm{\exists }{H}_{n_h}$ just like before, where $n_h\in \mathbb{N}$. Now notice that since $H_{n_h}$ is itself a subgroup then $h^{-1}\in H_{n_h}$. Thus then $h^{-1}\in H$ by definition, showing closure of the inverses.

Therefore, $H$ is a subgroup of $G$.

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