Math4302 Modern Algebra (Lecture 6)

$(\mathbb{Z}_n,+)$ is cyclic and $\mathbb{Z}_n=\langle 1\rangle=\{0,1,2,\cdots,n-1\}$

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$(\mathbb{Z},+)$ is cyclic and $\mathbb{Z}=\langle 1\rangle=\langle -1 \rangle$

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$S_3$ is not cyclic

$\langle e\rangle=\{e\}$ $\langle (1,2)\rangle=\{e,(1,2)\}$ $\langle (1,3)\rangle=\{e,(1,3)\}$ $\langle (2,3)\rangle=\{e,(2,3)\}$ $\langle (1,2,3)\rangle=\{e,(1,2,3),(1,3,2)\}$ $\langle (1,3,2)\rangle=\{e,(1,3,2),(1,2,3)\}$

Let $G=\langle a\rangle$ be a cyclic group, then $\forall g_1,g_2\in G$ we have $g_1g_2=g_2g_1$ since $g_1g_2=a^k_1a^k_2=a^{k_1+k_2}$ and $g_2g_1=a^k_2a^k_1=a^{k_1+k_2}$

$5$ in $(\mathbb{Z},+)$ has infinite order.

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$5$ in $(\mathbb{Z}_{10},+)$ has order $2$.

$\langle 5\rangle=\{0,5\}$.

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$5$ in $(\mathbb{Z}_{6},+)$ has order $6$.

$\langle 5\rangle=\{0,5,4,3,2,1\}$.

There are 2 cases:

Case 1:

There is no positive $n$ such that $a^n=e$.

Then $a^i\neq a^j$ if $i\neq j, i,j\in \mathbb{N}$.

Reason: if $a^i=a^j$, then $a^{i-j}=e$.

Then the order of group is infinite.

Case 2:

There is a positive $n$ such that $a^n=e$.

Let $n$ be the smallest such positive integer. Then we claim $\langle a^n\rangle=\{e,a^1,a^2,\cdots,a^{n-1}\}$.

We claim they are all distinct.

Suppose not, then we can have $a^i=a^j$ for $i\neq j$, $0\leq i,j\leq n-1$.

Then $a^{i-j}=e$ but $i-j\leq n-1$. Therefore $n$ is not the smallest positive integer such that $a^n=e$.

Case 1:

If $|G|=\infty$, then we can map $G$ to $(\mathbb{Z},+)$, where $G=\langle a\rangle$. $\phi(n)=a^n$. This gives a bijection between $G$ and $(\mathbb{Z},+)$.

where $\phi(n+m)=a^{n+m}=a^n a^m=\phi(n)\phi(m)$.

Case 2:

If $|G|=n$, then we can map $G$ to $(\mathbb{Z}_n,+)$, where $G=\langle a\rangle$. $\phi(n)=a^n$. This gives a bijection between $G$ and $(\mathbb{Z}_n,+)$.

where $\phi(n+m)=\phi(r)=a^{n+m}=a^n a^m=\phi(n)\phi(m)$.

Let $H=\langle (12)(345)\rangle\subseteq S_5$. Then $H\simeq \mathbb{Z}_6^+$.

Let $\tau=(12)(345)$

All the elements of $H$ are:

• $\tau^0=(12)(345)$
• $\tau^1=(453)$
• $\tau^2=(12)(534)$
• $\tau^3=(345)$
• $\tau^4=(12)(453)$
• $\tau^5=(534)$