1. For each of the following groups $$ G $$, determine whether $$ H $$ is a normal subgroup of G. If $$ H $$ is a normal subgroup, write out a Cayley table for the factor group $$ G / H $$. a. $$ G=S_{4} $$ and $$ H=A_{4} $$ b. $$ G=A_{5} $$ and $$ H=\{(1),(123),(132)\} $$ c. $$ G=S_{4} $$ and $$ H=D_{4} $$ d. $$ G=Q_{8} $$ and $$ H=\{1,-1, I,-I\} $$ e. $$ G=\mathbb{Z} $$ and $$ H=5 \mathbb{Z} $$

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a. Since $$ A_4 $$ is the alternating subgroup of $$ S_4 $$, and for any $$ n\ge5 $$ the alternating subgroup is normal in $$ S_n $$, while for $$ n=4 $$ one may check by conjugation that
$$
gA_4g^{-1}=A_4\quad\text{for all }g\in S_4,
$$
we conclude that $$ A_4 $$ is normal in $$ S_4 $$. Moreover, the index is
$$
[S_4:A_4]=\frac{24}{12}=2.
$$
Thus, the factor group $$ S_4/A_4 $$ is of order 2 (isomorphic to $$\mathbb{Z}_2$$). Its Cayley table is given by:

| $$*$$       | $$ A_4 $$         | $$ gA_4 $$ (any odd permutation coset) |
|-------------|-------------------|-----------------------------------------|
| $$ A_4 $$   | $$ A_4 $$         | $$ gA_4 $$                             |
| $$ gA_4 $$  | $$ gA_4 $$        | $$ A_4 $$                              |

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b. The subgroup
$$
H=\{(1), (123), (132)\}
$$
of $$ A_5 $$ has order 3. Since $$ A_5 $$ is a simple group (its only normal subgroups are the trivial subgroup and $$ A_5 $$ itself), no non‐trivial proper subgroup can be normal. Hence, $$ H $$ is not normal in $$ A_5 $$.

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c. Here
$$
G=S_4\quad\text{and}\quad H=D_4,
$$
where $$ D_4 $$ (of order 8) is one of the subgroups of $$ S_4 $$ isomorphic to the symmetry group of a square. Since $$ S_4 $$ has exactly one nontrivial proper normal subgroup (namely $$ A_4 $$) and because $$ D_4 $$ is not contained in $$ A_4 $$ (its order 8 does not divide 12), it follows by conjugation arguments that $$ D_4 $$ is not normal in $$ S_4 $$.

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d. The quaternion group is
$$
Q_8=\{1,-1,I,-I,J,-J,K,-K\},
$$
and the subgroup under consideration is
$$
H=\{1,-1,I,-I\}.
$$
It is a known fact that every subgroup of $$ Q_8 $$ is normal (in fact, $$ Q_8 $$ is a Hamiltonian group). The order of $$ H $$ is 4, so the factor group $$ Q_8/H $$ has order
$$
|Q_8/H|=\frac{8}{4}=2.
$$
Thus, $$ Q_8/H $$ is isomorphic to $$\mathbb{Z}_2$$ and its Cayley table is:

| $$*$$           | $$ H $$      | $$ JH $$ (choose any coset representative outside $$ H $$) |
|-----------------|--------------|--------------------------------------------------------------|
| $$ H $$         | $$ H $$      | $$ JH $$                                                     |
| $$ JH $$        | $$ JH $$     | $$ H $$                                                      |

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e. The group $$ \mathbb{Z} $$ is abelian, so every subgroup is normal. The subgroup $$ H=5\mathbb{Z} $$ consists of all multiples of 5, and the factor group $$ \mathbb{Z}/5\mathbb{Z} $$ is a cyclic group of order 5 (isomorphic to $$ \mathbb{Z}_5 $$). Its Cayley table (using representatives $$ 0, 1, 2, 3, 4 $$) is:

| $$+$$   | $$0$$ | $$1$$ | $$2$$ | $$3$$ | $$4$$ |
|---------|-------|-------|-------|-------|-------|
| **$$0$$** | $$0$$ | $$1$$ | $$2$$ | $$3$$ | $$4$$ |
| **$$1$$** | $$1$$ | $$2$$ | $$3$$ | $$4$$ | $$0$$ |
| **$$2$$** | $$2$$ | $$3$$ | $$4$$ | $$0$$ | $$1$$ |
| **$$3$$** | $$3$$ | $$4$$ | $$0$$ | $$1$$ | $$2$$ |
| **$$4$$** | $$4$$ | $$0$$ | $$1$$ | $$2$$ | $$3$$ |
a. $$ A_4 $$ is a normal subgroup of $$ S_4 $$. The factor group $$ S_4 / A_4 $$ has two elements and is isomorphic to $$ \mathbb{Z}_2 $$. b. $$ H = \{(1), (123), (132)\} $$ is not a normal subgroup of $$ A_5 $$. c. $$ D_4 $$ is not a normal subgroup of $$ S_4 $$. d. $$ H = \{1, -1, I, -I\} $$ is a normal subgroup of $$ Q_8 $$. The factor group $$ Q_8 / H $$ is isomorphic to $$ \mathbb{Z}_2 $$. e. $$ H = 5\mathbb{Z} $$ is a normal subgroup of $$ \mathbb{Z} $$. The factor group $$ \mathbb{Z} / 5\mathbb{Z} $$ is isomorphic to $$ \mathbb{Z}_5 $$.

For each of the following groups , determine whether is a normal subgroup of
G. If is a normal subgroup, write out a Cayley table for the factor group .
a. and
b. and
c. and
d. and
e. and

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