Math4302 Modern Algebra (Lecture 5)

For an arbitrary group $(G,*)$,

$(\{e\},*)$ and $(G,*)$ are always subgroups.

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$(\mathbb{Z},+)$ is a subgroup of $(\mathbb{R},+)$.

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Non-example:

$(\mathbb{Z}_+,+)$ is not a subgroup of $(\mathbb{Z},+)$.

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Subgroup of $\mathbb{Z}_4$:

$(\{0,1,2,3\},+)$ (if $1\in H$, $3\in H$)

$(\{0,2\},+)$

$(\{0\},+)$

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Subgroup of $\mathbb{Z}_5$:

$(\{0,1,2,3,4\},+)$

$(\{0\},+)$

Cyclic group with prime order has only two subgroups

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Let $D_n$ denote the group of symmetries of a regular $n$-gon. (keep adjacent points pairs).

$$D_n=\{\sigma\in S_n\mid i,j\text{ are adjacent } \iff \sigma(i),\sigma(j)\text{ are adjacent }\}$$ $$\begin{pmatrix} 1&2&3&4\\ 2&3&1&4 \end{pmatrix}\notin D_4$$

$D_4$ has order $8$ and $S_4$ has order $24$.

$|D_n|=2n$. ($n$ option to rotation, $n$ option to reflection. For $\sigma(1)$ we have $n$ option, $\sigma(2)$ has 2 option where the remaining only has 1 option.)

Since $1-4$ is not adjacent in such permutation.

$D_n\leq S_n$ ($S_n$ is the symmetric group of $n$ elements).

If $H$ is subgroup, then $e\in H$, so $H$ is non-empty and if $a,b\in H$, then $b^{-1}\in H$, so $ab^{-1}\in H$.

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If $H$ has the given property, then $H$ is non-empty and if $a,b\in H$, then $ab^-1\in H$, so

• There is some $a,a\in H$, $aa^{-1}\in H$, so $e\in H$.
• If $b\in H$, then $e\in H$, so $eb^{-1}\in H$, so $b^{-1}\in H$.
• If $b,c\in H$, then $c^{-1}$, so bc^{-1}^{-1}\in H, so $bc\in H$.