"Abstract algebra" 03-quotient group and direct product

1. Companion Set

Now continue to study the decomposition of the group, first to discuss the relationship between the General subgroup, as well as the subgroups and the parent group. First, depending on the criteria of the subgroup, if \ (H,k\leqslant g\), it is easy to have \ (H\cap g\leqslant g\). So \ (H\cup g\)? Of course, here \ (h,k\) are true subgroups and do not contain each other. Taking elements \ (h\not\in k\) from \ (h\), taking elements from \ (k\) (K\not\in h\), it is easy to prove \ (hk\not\in H\cup g\), so \ (h,k\) must not be a subset of \ (g\).

What if \ (Hk\) is included in it, that is \ (hk\) is a subset of \ (g\)? For \ (H_1k_1,h_2k_2\in hk\), if there is always a \ ((h_1k_1) (h_2k_2) =hk\), it is easy to prove that the condition and \ (hk=kh\) are equivalent. So there is the following conclusion, but note that \ (hk=hk\) does not mean \ (hk=kh\). Such a partition requires a subset to meet certain conditions, does not meet our current needs, need to find another method.

\[hk\leqslant G\quad\leftrightarrow\quad Hk=kh\tag{1}\]

Now it seems that we have to abandon the idea of decomposing the parent into several subgroups, but only a subgroup \ (h\) as a reference or unit. We also want to divide each chunk as large as the subgroup, and the best elements have a one by one correspondence with \ (h\). Thus we think of the set of Ah\, which represents the sum of the products of each element of \ (a\) and \ (h\), which is called the left companion set of \ (h\), and \ (coset) is the representative element of the left companion set. If \ (A\in h\), obviously \ (ah=h\), now comes to study \ (A\not\in h\) when the relationship between \ (ah\).

for arbitrary \ (B\in ah\), exists \ (B=ah, (h\in h) \), then \ (bh=ahh=ah\), that is, any element in \ (ah\) is a left-hand set representing the meta, which is fully coincident with \ (ah\). In other words, all left-side sets are either completely equal or have no intersection, and each element is divided into a left-hand set, and can be used as a representation of the left-hand set. On the other hand, to \ (B\in ah\), there is \ (A^{-1}b=h\in h\), easy to prove that \ (a^{-1}b\in h\) is \ (a,b\) belong to a left-hand set of necessary and sufficient conditions, it is an equivalence relationship between the group elements.

Also can define right accompany set \ (ha\) concept, and have and left accompany set the same conclusion, but belong to a right accompany set condition to change to \ (ab^{-1}\). For non-commutative groups, \ (ah\) and \ (ha\) are not generally equal, so the separation of the left and right companion sets is completely different (\ (h\) itself, except that it is both left-hand set, and is set to the right. But you may not be reconciled, there must be another way to connect them. Considering that the left and right to accompany set is just upside-down, you can naturally think of inverse, to any \ (Ah\in ah\), there is a ((AH) ^{-1}=h^{-1}a^{-1}\in ha^{-1}\). The elements of \ (ah\) and \ (ha^{-1}\) are completely reciprocal, so that the left and right companion sets find one by one corresponding relationships. Now want to, left accompany set \ (ah\) in the inverse of the elements are scattered to the other left to accompany the concentration, but magically concentrated in the right to accompany set \ (ha^{-1}\).

Consider the set of all left-hand sets \ (ah\), whose order is called a subset \ (h\) index , recorded as \ ([g:h]\), then apparently the formula (2) of the Lagrange theorem is established. Further, if \ (k\leqslant h\leqslant g\), it is also easy to have the formula (3) established (pay attention to the endless discussion). And it can be intuitively seen, \ (K\) The accompanying set is in the \ (h\) accompany set on the basis of \ (K\) for the division of the unit.

\[| g|=| h| [G:h]\tag{2}\]

\[[g:k]=[g:h][h:k]\tag{3}\]

Now see \ (H\cap k\) accompany set and \ (H,k\) with the relationship between the set, first by the conclusion is known, \ (H\cap k\) accompany set is just \ (h,k\) accompany set of a again split. thus \ (Ah\cap bk\) is either an empty set, or just a certain \ (H\cap k\) companion sets. Further, if \ (C=ah\cap bk\), then \ (Ah\cap bk=ch\cap ck=c (h\cap K) \), i.e. \ (Ah\cap bk\) contains only one \ (H\cap k\) companion set. In this case, it is easy to have the following inequalities.

\[[g:h\cap k]\leqslant [g:h][g:k]\tag{4}\]

Finally, the subset \ (hk\), which is obviously composed of some \ (K\) left companion set. Also consider \ (h\) in \ (H\cap k\) \ (m=\dfrac{| h|} {| H\cap k|} \) A left accompany set, considering \ (H_1,h_2\in h\), they belong to the same \ (H\cap k\) left to accompany set the necessary and sufficient conditions are \ (h_1^{-1}h_2\in h\cap k\). And this condition is obviously equivalent to \ (H_1^{-1}h_2\in k\), it is \ (h_1,h_2\) belong to the same \ (K\) the left accompany set yell condition, so \ (hk\) in \ (K\) The number of left accompany set is \ (m\). The above conclusions can be summed up as formula (5), obviously only when \ (H\cap k={e}\), only \ (| hk|=| h| | k|\).

\[| hk|=\dfrac{| h| | k|} {| H\cap k|} \tag{5}\]

2. Homomorphism and quotient group 2.1 homomorphism theorem

Now the group \ (g\) is divided into \ (h\) the companion set, \ (h\) Of course there is a finer division method, now need to study its companion set composed of the collection. Instead of directly studying the companion set, we adopt a more general approach. Recalling the definition of a companion set is actually a mapping from the group element to the companion set, and we want to study the mapping between the general algebraic systems. Considering the mapping between the two systems \ (\langle s,\circ\rangle, \langle \bar{s},\star\rangle \) \ (f\), we certainly hope that the operation law is maintained, and that the mappings that meet the following conditions are called homomorphic mappings (homomorphism). If the mapping is full, it is called \ (s,\bar{s}\) homomorphism, which is recorded as \ (s\sim\bar{s}\).

\[f (A\circ B) =f (a) \star f (b) \tag{6}\]

Our main concern is of course the relationship between homomorphic mapping and the original image, that is, the relationship between homomorphic systems. If \ (g\sim\bar{g}\), where \ (g\) is a group, it is easy to prove \ (\bar{g}\) satisfies all the conditions of the group (as an exercise), so \ (\bar{g}\) is also a group. Of course, you can also get more conclusions, such as unit element mapping to the unit, inverse mapping to the inverse, and even sub-group mapping to subgroups, here do not repeat. Instead of thinking about homomorphic mappings, each of its images may have more than one image, and \ (g\) is divided into different equivalence classes according to the difference of the image, what are the properties of these equivalence classes? What does it have to do with \ (\bar{g}\)?

Obviously those equivalence classes correspond to the homomorphism like one by one, and if they can be defined, they are naturally isomorphic, and the task now is to look for meaningful operations of these equivalence classes. The first definition \ (\bar{e}\) of the original image \ (F^{-1} (\bar{e}) is the kernel (kernel), recorded as \ (\text{ker}\:f\). Let's look at what those equivalence classes are, for \ (\bar{x}\in\bar{g}\), investigate \ (X=f^{-1} (\bar{x}) \). For any \ (a,b\in x\), \ (f (a^{-1}b) = (f (a)) ^{-1}f (b) =\bar{e}\), so \ (A^{-1}b\in \text{ker}\:f\), thus \ (k=\text{ker}\:f\) is a subgroup, And each equivalence class is its left companion set. You can also find that the evidence just right with the set is also set up, that is, \ (\text{ker}\:f\) of the set is the same!

Since the companion set is not divided about, it can be defined by the \ (Ak\cdot bk= (AB) k\), it is easy to prove that in this operation, \ (K\) The accompanying set and \ (\bar{g}\) is isomorphic. We need to specialize in sub-groups such as the nucleus (n\), which requires \ (An=na\) to be established. For this definition, the subgroup that satisfies the following formula is a normal subgroup (normal subset), recorded as \ (N\trianglelefteq g\), if \ (N\neq g\), also recorded as \ (N\triangleleft g\). Just now the conclusion can be said, the nucleus of homomorphic mapping is normal subgroup, then in turn? In fact, it is easy to prove that, for any normal subgroup \ (n\), mapping \ (f (a) =an\) is the same state. So we can conclude that any normal subgroup is equivalent to a homomorphism mapping.

\[\forall a\in G (ana^{-1}) \quad\rightarrow\quad N\trianglelefteq g\tag{7}\]

Because the companion set of the normal subgroup \ (n\) corresponds to homomorphic image one by one, they are bound to form a group, which is defined as quotient group (quotient group), which is recorded as \ (g/n\), thus having \ (| g/n|=[g:n]\). The conclusion just now is that the symbol is the next type, which is called the homomorphism basic theorem.

\[g\sim G/n\cong \bar{g}\tag{8}\]

Now we continue to make some regular discussions about normal subgroups. The normal subgroup is the relationship between \ (n\) and \ (g\), so to arbitrary \ (n\leqslant h\leqslant g,n\trianglelefteq g\), there is always \ (N\trianglelefteq h\), but to \ (h\ Trianglelefteq N\trianglelefteq g\), but not necessarily a \ (H\trianglelefteq g\). The subgroups of the commutative group are obviously normal subgroups. For non-commutative groups \ (g\), \ (\{e\}\) and \ (g\) are obviously normal subgroups, but if there are no other normal subgroups other than these two ordinary normal subgroups, then \ (g\) is called simple groups (single group). Conversely, if all its subgroups are normal subgroups, it is also called the Hamiltonian group . It is easy to prove that the intersection and product of two normal subgroups is also a normal subgroup (formula (8)), but the intersection and product of normal subgroups and subgroups can only be normal subgroups.

\[n,k\trianglelefteq g\quad\rightarrow\quad n\cap k\trianglelefteq g,\quad Nk\trianglelefteq G\tag{8}\]

Consider several questions about the normal subgroup:

　　? \ (a_n\) is the normal subgroup of \ (s_n\), \ (k_4\) is the normal subgroup of \ (s_4\). If \ (N\neq 4\) is known, \ (a_n\) is simple groups, then the non-trivial normal subgroup of \ (s_n\) is only \ (a_n\);

　　? \ (N,k\trianglelefteq g\) and \ ((| n|,| k|) =1\), if \ (g/n,g/k\) are commutative group, verification \ (g\) is also the Exchange Group;

　　? \ (N=\langle a\rangle\) is a normal subgroup, then any \ (H\leqslant n\) is also a normal subgroup;

In this paper, a method of analyzing the structure of a group is given, which divides the group into normal subgroups and quotient groups, and introduces the isomorphism theorem of the well-known group. The first isomorphism theorem is actually the homomorphism fundamental theorem, and the third isomorphism theorem takes the normal subgroup \ (n\) as the unit element, and obtains the structure of the larger normal subgroups. The second isomorphism theorem is obtained by replacing \ (g\) with \ (hn\).

(1) First isomorphism theorem: \ (g/\text{ker}\:f\cong f (G) \);

(2) Second isomorphism theorem: \ (n\trianglelefteq g,\:h\leqslant g\quad\rightarrow\quad hn/n\cong h/(h\cap N) \);

(3) Third isomorphism theorem: \ (h,n\trianglelefteq g,\:n\subseteq h\quad\rightarrow\quad g/h\cong (g/n)/(h/n) \).

2.2 Self-isomorphic group

In the previous chapter, the symmetry group, whose constituent elements are the one by one mappings of a set, is now viewed as a special subgroup on the group. We consider a set of all the automorphism transformations of a group, it is easy to prove that they are composed of groups, called automorphism groups (automorphism), and are written as \ (\text{aut}\:g\). It is easy to prove that the automorphism group of the infinite cyclic group is the \ (2\) Order Cycle Group, while the automorphism group of the \ (n\) Order cyclic group is the \ (\varphi (n) \) Order group. If you feel that the automorphism group is not well constructed, then you can try the isomorphic mapping \ (\sigma_a (x) \to axa^{-1}\), all of which constitute an inner automorphism group , which is recorded as \ (\text{inn}\:g\). It is obvious that normal subgroups remain unchanged under the isomorphism, so it is also called invariant subgroup . It is also easy to prove that the inner automorphism group is the normal subgroup of the automorphism group (equation (9)).

\[\text{inn}\:g\trianglelefteq\text{aut}\:g\tag{9}\]

Now consider the mapping from \ (g\) to \ (\text{inn}\:g\), which is obviously homomorphic, and the kernel of the map is all the \ (a\) (internal automorphism unit element is the identity transform) (axa^{-1}=x\). For this we define the elements that are interchangeable with all elements as the central element, the subgroup of which is called the Center (center), which is remembered as \ (C (G) \) or \ (c\), and the center has only \ (\{e\}\) of the group called the No-central group. In this way, the use of isomorphism theorem is the following form.

\[\text{inn}\:g\cong G/c\tag{10}\]

For the general automorphism group, there is no significant result, and there is no special relationship between it and the original group. If the automorphism group \ (\text{aut}\:g\) has a center, take a non-identical automorphism transform \ (\tau (a) =b\neq a\) and inner automorphism \ (\sigma_a\), thus having \ (\tau\sigma_a=\sigma_a\tau\), You can then prove that \ (a^{-1}b\) is the center of \ (g\). Thus if \ (\text{aut}\:g\) has a center, then \ (g\) also has the center. Conversely, if \ (g\) has no center, then \ (\text{aut}\:g\) has no center. Consider the following questions:

　　? Proof \ (s_n\) and \ (\text{aut}\:s_n\) are no central groups;

　　? The necessary and sufficient condition to prove that the automorphism group of the (n\) Order cyclic group is a cyclic group is \ (n=2,4,p^e,2p^e\), where \ (p\) is an odd prime.

3. Direct product

Normal subgroups can decompose groups into two groups, but these two groups are not at the same level and do not seem to be "decomposed" in real sense. Our ideal decomposition should be that each part is independent of each other and is Order independent, it is like being divided into different dimensions. For this reason, we first construct a kind of group satisfying the condition, to the group \ (a_1,a_2,\cdots,a_n\), examine the following set \ (g\). In \ (g\) The multiplication is defined ((A_1,\cdots,a_n) (b_1,\cdots,b_n) = (a_1b_1,\cdots,a_nb_n) \), it is easy to prove that under this multiplication, \ (g\) is a group. If you make a subset \ (\{e_1,\cdots,x_k,\cdots,e_n\}\) (g_k\), obviously \ (G_k\) is a group that is isomorphic to \ (a_k\).

\[g=a_1\times a_2\times\cdots\times a_n=\{(x_1,x_2,\cdots,x_n) \}\tag{11}\]

For the above \ (g\) decomposition \ (g_k\) Obviously meet our needs: (1) \ (g_k\) are normal subgroups; (2) \ (G=g_1g_2\cdots g_n\); (3) \ (g_1\cdots g_{k-1}\cap g_k=\{e\}\). More fundamentally, it satisfies our requirements for independent decomposition: Each part is independent and order independent, and the mathematical language description is the following equivalence condition (proof as exercise). The decomposition that satisfies the above definition or the following equivalence condition is called \ (g\) The Direct product (direct product), and does not confuse the case also writing \ (G=g_1\times g_2\times\cdots\times g_n\).

(1) \ (g=g_1g_2\cdots g_n\) The decomposition of the existence and unique, wherein \ (g\in g,g_k\in g_k\);

(2) \ (g_i,g_j\) The elements are multiplied interchangeably, i.e. \ (g_ig_j=g_jg_i\) constant.

The straight integral solution decomposes the group into completely independent subgroups, which facilitates further study, the group which can do the straight integral solution is generally called the decomposition group , if the decomposed subgroups are simple groups, it is called the fully decomposed group . We have two basic questions: What kind of group can be decomposed? Can normal subgroups be used as decomposition factors? The answer to the first question is not easy, and we can only judge a few simple scenarios at the moment. For example, for the cyclic group, it can be proved that the subgroups of the infinite cyclic group and the finite cycle group of the Order prime power have a common part, and the ancient capital is non-decomposed. For a cyclic group with multiple elements of order (G=\langle a\rangle\), the order has a mutual element decomposition (M=m_1m_2\cdots m_n\), and the knowledge of elementary number theory can have the following straight integral solutions.

\[g=\langle A^{\frac{m}{m_1}}\rangle\langle A^{\frac{m}{m_2}}\rangle\cdots\langle a^{\frac{m}{m_n}}\rangle,\quad\ Left|\langle A^{\frac{m}{m_k}}\rangle\right|=m_k\tag{12}\]

So is it possible for each normal subgroup to be a decomposition factor? This is actually true for the fully decomposed group. Imagine if \ (G=g_1\times g_2\times\cdots\times g_n\) is a complete decomposition and has \ (N\trianglelefteq g\). First there's \ (N\cap G_k\trianglelefteq g_k\), and \ (g_k\) is simple groups, so there \ (N\cap g_k=g_k\) or \ (N\cap g_k=\{e\}\). This means \ (n\) falls completely in some \ (g_k\), it must be some \ (g_k\) of the direct product, so it can also be used as a decomposition factor. In addition, you can prove that if \ (G=n\times k\), then \ (G/n\cong k\), it pulls the quotient to a position parallel to \ (n\).

One more question to consider is how the subgroups in the decomposed group are decomposed. If \ (g=g_1\times g_2\times\cdots\times g_n\), we hope that the next type can be set up, but it is necessary to establish the conditions. The necessary and sufficient condition to prove its existence is \ (| g_k|\) coprime, the full use of the cycle group has just been discussed proof \ (a\) decomposition of each factor is its generation of the elements of the group, the necessity is through the construction of two \ (p-\) Order (refer to the next article) of the product to export contradictions. In addition, if \ (G=g_1\times g_2\) and \ (G_1\leqslant h\), it is easy to prove that there is \ (H=g_1\times (G_2\cap H) \).

\[h= (H\cap g_1) \times (H\cap g_2) \times\cdots (H\cap g_n) \tag{13}\]

"Abstract algebra" 03-quotient group and direct product