Math 361, Spring 2014, Assignment 8
Carefully define the following terms, then give one example and one non-example of each:[edit]
- Meet (of two subgroups of a group $G$).
- Join (of two subgroups).
- Product (of two subgroups).
- Subnormal series.
- Factors (of a subnormal series).
- Isomorphic (subnormal series).
- Refinement (of a subnormal series).
- Saturated (subnormal series).
- Composition series.
- Composition factors.
Carefully state the following theorems (you need not prove them):[edit]
- Lemma relating joins to products.
- First isomorphism theorem.
- Second isomorphism theorem.
- Third isomorphism theorem.
- Jordan-Hölder Theorem.
Solve the following problems:[edit]
- Section 34, problems 3 and 5.
- Section 35, problems 1, 3, and 7.
Questions:[edit]
Solutions:[edit]
Definitions:[edit]
- Meet (of two subgroups of a group $G$).
Let $H,K$ be subgroups of $G$.
The meet of $H$ and $K$, written $H\wedge K$, is the intersection $H\cap K$. This is again a subgroup of $G$, and $H\wedge K$ is the largest subgroup contained in both $H$ and $K$.
Example:
In $\mathbb{Z}_{12}$: $$\langle 4\rangle\wedge\langle 6\rangle = \langle 0\rangle$$ $$\langle 2\rangle\wedge\langle 3\rangle = \langle 6\rangle$$ $$\langle 3\rangle\wedge\langle 4\rangle = \langle 0\rangle$$
Non-example:
$$\langle 4\rangle\wedge\langle 6\rangle\neq\langle 2\rangle$$
- Join (of two subgroups).
The join of $H$ and $K$, written $H\vee K$, is the subgroup generated by $H\cup K$. Observation: $H\vee K$ is the smallest subgroup contained in both.
Example:
In $\mathbb{Z}_{12}$: $$\langle 4\rangle\vee\langle 6\rangle = \langle 2\rangle$$ $$\langle 2\rangle\vee\langle 3\rangle = \langle 1\rangle$$
Non-example:
$$\langle 4\rangle\vee\langle 6\rangle\neq\langle 0\rangle$$
- Product (of two subgroups).
The product of $H$ and $K$, written $HK$, is given by: $$HK=\lbrace hk | h\in H, k\in K\rbrace$$.
Note: If either H or K is normal, then the product is a subgroup and $HK=H\vee K$.
- Subnormal series.
Let $G$ be a group. A subnormal series for $G$ is a chain of normal subgroups, denoted: $$H_1\unlhd H_2\unlhd H_3\unlhd\cdots\unlhd H_n$$.
- Factors (of a subnormal series).
The quotient groups $H_i / H_{i-1}$ are called the factors of the series.
- Isomorphic (subnormal series).
Two subnormal series: $$H_0\unlhd\cdots\unlhd H_n$$ $$K_0\unlhd\cdots\unlhd K_m$$ are isomorphic if they have isomorphic factors.
Example:
In $\mathbb{Z}_{12}$: $$\langle 0\rangle\subseteq\langle 4\rangle\subseteq\langle 2\rangle\subseteq\mathbb{Z}_{12}$$ $$\langle 0\rangle\subseteq\langle 6\rangle\subseteq\langle 3\rangle\subseteq\mathbb{Z}_{12}$$ are isomorphic.
- Refinement (of a subnormal series).
A refinement of a subnormal series is any other subnormal series obtained by inserting new subgroups into the chain.
Example
In $\mathbb{Z}_{12}$: $$\langle 0\rangle\subseteq\langle 4\rangle\subseteq\langle 2\rangle\subseteq\mathbb{Z}_{12}$$ is a refinement of $$\langle 0\rangle\subseteq\langle 2\rangle\subseteq\mathbb{Z}_{12}$$
- Saturated (subnormal series).
A subnormal series is saturated if it has no refinements.
Example
In $\mathbb{Z}_{12}$: $$\langle 0\rangle\subseteq\langle 4\rangle\subseteq\langle 2\rangle\subseteq\mathbb{Z}_{12}$$
Non-example
In $\mathbb{Z}_{12}$: $$\langle 0\rangle\subseteq\langle 2\rangle\subseteq\mathbb{Z}_{12}$$
- Composition series.
A composition series for $G$ is a saturated subnormal series.
Example
In $\mathbb{Z}_{12}$: $$\langle 0\rangle\subseteq\langle 6\rangle\subseteq\langle 2\rangle\subseteq\mathbb{Z}_{12}$$
Non-example
In $\mathbb{Z}_{12}$: $$\langle 0\rangle\subseteq\langle 3\rangle\subseteq\mathbb{Z}_{12}$$
- Composition factors.
The composition factors of $G$ are the factors of any composition series.
Example:
The composition factors of $\mathbb{Z}_{12}$ are $\lbrace\mathbb{Z}_2,\mathbb{Z}_2,\mathbb{Z}_3\rbrace$.
Non-example:
The composition factors of $\mathbb{Z}_{12}$ are not$\lbrace\mathbb{Z}_2,\mathbb{Z}_3\rbrace$, you need to show the repetition of the $\mathbb{Z}_2$ factor.
Theorems:[edit]
- Lemma relating joins to products.
If either $H$ or $K$ is normal, then $HK$ is a subgroup and $HK= H\vee K$
- First isomorphism theorem.
aka: The fundamental theorem of homomorphisms
Suppose $\phi :G\rightarrow G'$ is a homomorphism. Then there exists a unique monomorphism $\phi^* :G/ker\phi \rightarrow G'$. In particular $im\phi\simeq G/ker\phi$.
- Second isomorphism theorem.
Suppose $H$ and $K$ are subgroups of $G$ and $K$ is normal. Then: $$\frac{HK}{K}\simeq\frac{H}{H\wedge K}$$
- Third isomorphism theorem.
Suppose $K\subseteq H\subseteq G$ and both $H$ and $K$ are normal in $G$. Then, $$\frac{G/K}{H/K}\simeq\frac{G}{H}$$
- Jordan-Hölder Theorem.
Any two composition series for $G$ are isomorphic.