Why do authors usually mix the "isomorphic" and "equal" on sign?
A global use of the "$A=B$" is to denote that object $A$ is identical with object $B$. In the set theory this means that $A$ and $B$ have the same elements. But, frequently in some of Papers about Algebraic-Structure we see some propositions like the following:
1. [Abelian p-group $H_p$ as the product of p-cyclic group]
\[H_p = (\mathbb{Z}/p^{e_1}\mathbb{Z}) \times (\mathbb{Z}/p^{e_2}\mathbb{Z}) \times \cdots \times (\mathbb{Z}/p^{e_r}\mathbb{Z}) \]
2. [Projectional result Linear Functional $f:A \rightarrow \mathbb{C}$, where $A$ is $n$-dimensional vector space for $n\geq 2$]
\[A = \text{ker} (f) \oplus \mathbb{C}\]
Those propositions don't make any sense if we think "=" as equal sign. For the first proposition implies that any abelian p-group no matter how strange its elements is, can be described as the well-known product of modulo $p^{e_i}$ integers.
The second proposition is even more confusing, since $A$ can be any abstract vector space and hence $\text{ker} (f)$ is also an abstract vector space, we can't even "add" the element of $A$ with the element of $\mathbb{C}$.
We can think the "=" as isomorphic on both propositions and the problem is solved, but occasionally in the same paper/book the Authors use "$\cong$" and they often mix it with "$=$". Depending on the focus of the paper or book this might not be a serious problem, for example in the paper about group structure it is safe to think isomorphic as "the same".
Consider you are being asked by someone who bother the difference between "$=$" and "$\cong$", he asks you about the statement of Fundamental Theorem of Finite Abelian Group which states that :
Every Finite Abelian Group is the direct product of its cyclic subgroups.
Suppose he tested you with a challenge, he gave you a finite abelian group $G$, and want you to verified the statement of above theorem.
Suppose that you're given a finite abelian group $G$ and then you want to find the sets $A_1,A_2,\cdots,A_k$ which are cyclic subgroups of $G$, so that every $x \in G$ satisfies $x= a_1 a_2 \cdots a_k$ where $a_i \in A_i$. Yes Of Course, by Fundamental Theorem of Finite Abelian Group we can always do this,
for example if $G=\{e,a,b,ab\}$ where $a^2=b^2=e$ then we can take $A_1 = \{e, a\}$, $A_2\{e,b\}$ and $A_3=\{e,ab\}$, therefore $A_1 A_2 A_3 = G$.
Note that here we take an internal product, since for the case of $G$ is abelian group every subgroup of $G$ is a normal subgroup, hence the external direct product implies the internal direct product (note that $A_i \cap A_j = \{e\}$).
We can also "cheat the someone" since he/she doesn't give us a proper form of $G$, we can start the "cheat" by first stating that:
"I let $G=\{(0,0,0), (a_1,b_1,c_1), (a_2,b_2,c_2),\cdots, (a_n,b_n,c_n) \}$, since you don't give me the exact form of $G$, I must name its elements by something, and I choose to name them by these variables". And then, we pretend that we first know nothing about $a_i,b_i$, and $c_i$ (that we just want to name them that way) and it turns out that it is not our fault then
$\{(0,0,0),(a_1,0,0) ,(a_2,0,0),\cdots,(a_n,0,0)\}$,
$\{(0,0,0), (0,b_1,0),(0,b_2,0),\cdots,(0,b_n,0)\}$ and
$\{(0,0,0),(0,0,c_1),(0,0,c_2),\cdots,(0,0,c_n)\}$
are all become cyclic group :D (Of course we already know that they will).
But according to proposition 1, $\mathbb{Z}/p^{e_i} \mathbb{Z}$ is not always a subgroup of $H_p$, they have different elements. We still can resolve it like when we "cheat"above by stating that the elements of $H_p$ are $(h_1,\cdots,h_r)$.
Suppose that someone is one step ahead, he/she asks you by first giving the exact element of $H_p$, say $H_p$ is abelian group of order $8$ given by $\{e,a_2,\cdots,a_8\}$, to make it a non-cyclic group (after some calculations or cayley table) $H_p=\{e, a_1, a_2, a_1a_2, b_1,a_1b_1, a_2b_1, a_1a_2b_1\}$ and $a_2^2=e$, $a_1^2=a_2$, $b_1^2=e$, and $a_i b_j = b_j a_i$ (he/she should agree with these, since this is an effort to make $H_p$ a non-cyclic abelian group of order 8, yes there might be other way). Since $a_1^2=a_2$, $H_p$ depends only on $a_1$ and $b_1$, so we can name $a=a_1$ and $b=b_1$ and hence
\[H_p = \{e,a,a^2,a^3,b,ab,a^2b\}\]
Do you realize? we've just stated
$$\{e,a_2,\cdots,a_8\}= \{e, a_1, a_2, a_1a_2, b_1,a_1b_1, a_2b_1, a_1a_2b_1\} = \{e,a,a^2,a^3,b,ab,a^2b\}$$
and we seem okay, though we throw $a_8$, $a_4$, $b_2$ etc. And we feel fine about that. We stated the three sets are the same, even though they are virtually not have the same elements. That's because we agree $a_1$ is $a$ and $b_1$ is $b$, and $a_2$ is $a_1^2$ and so on..
We agree! That is important in Mathematics.
Back on, what is actually "$=$" on proposition 1 and 2 ? In general it is isomorphism sign.
With some agreement, we can make it equal sign
(You know not everyone on this world use symbol $2$ to denote "two", in ancient roman its "II", in arabic it's different also, but are they really different in meaning? no, even sounded different in many languages, two is always two :D )
Hello formalism? :D :D
1. [Abelian p-group $H_p$ as the product of p-cyclic group]
\[H_p = (\mathbb{Z}/p^{e_1}\mathbb{Z}) \times (\mathbb{Z}/p^{e_2}\mathbb{Z}) \times \cdots \times (\mathbb{Z}/p^{e_r}\mathbb{Z}) \]
2. [Projectional result Linear Functional $f:A \rightarrow \mathbb{C}$, where $A$ is $n$-dimensional vector space for $n\geq 2$]
\[A = \text{ker} (f) \oplus \mathbb{C}\]
Those propositions don't make any sense if we think "=" as equal sign. For the first proposition implies that any abelian p-group no matter how strange its elements is, can be described as the well-known product of modulo $p^{e_i}$ integers.
The second proposition is even more confusing, since $A$ can be any abstract vector space and hence $\text{ker} (f)$ is also an abstract vector space, we can't even "add" the element of $A$ with the element of $\mathbb{C}$.
We can think the "=" as isomorphic on both propositions and the problem is solved, but occasionally in the same paper/book the Authors use "$\cong$" and they often mix it with "$=$". Depending on the focus of the paper or book this might not be a serious problem, for example in the paper about group structure it is safe to think isomorphic as "the same".
Consider you are being asked by someone who bother the difference between "$=$" and "$\cong$", he asks you about the statement of Fundamental Theorem of Finite Abelian Group which states that :
Every Finite Abelian Group is the direct product of its cyclic subgroups.
Suppose he tested you with a challenge, he gave you a finite abelian group $G$, and want you to verified the statement of above theorem.
Suppose that you're given a finite abelian group $G$ and then you want to find the sets $A_1,A_2,\cdots,A_k$ which are cyclic subgroups of $G$, so that every $x \in G$ satisfies $x= a_1 a_2 \cdots a_k$ where $a_i \in A_i$. Yes Of Course, by Fundamental Theorem of Finite Abelian Group we can always do this,
for example if $G=\{e,a,b,ab\}$ where $a^2=b^2=e$ then we can take $A_1 = \{e, a\}$, $A_2\{e,b\}$ and $A_3=\{e,ab\}$, therefore $A_1 A_2 A_3 = G$.
Note that here we take an internal product, since for the case of $G$ is abelian group every subgroup of $G$ is a normal subgroup, hence the external direct product implies the internal direct product (note that $A_i \cap A_j = \{e\}$).
We can also "cheat the someone" since he/she doesn't give us a proper form of $G$, we can start the "cheat" by first stating that:
"I let $G=\{(0,0,0), (a_1,b_1,c_1), (a_2,b_2,c_2),\cdots, (a_n,b_n,c_n) \}$, since you don't give me the exact form of $G$, I must name its elements by something, and I choose to name them by these variables". And then, we pretend that we first know nothing about $a_i,b_i$, and $c_i$ (that we just want to name them that way) and it turns out that it is not our fault then
$\{(0,0,0),(a_1,0,0) ,(a_2,0,0),\cdots,(a_n,0,0)\}$,
$\{(0,0,0), (0,b_1,0),(0,b_2,0),\cdots,(0,b_n,0)\}$ and
$\{(0,0,0),(0,0,c_1),(0,0,c_2),\cdots,(0,0,c_n)\}$
are all become cyclic group :D (Of course we already know that they will).
But according to proposition 1, $\mathbb{Z}/p^{e_i} \mathbb{Z}$ is not always a subgroup of $H_p$, they have different elements. We still can resolve it like when we "cheat"above by stating that the elements of $H_p$ are $(h_1,\cdots,h_r)$.
Suppose that someone is one step ahead, he/she asks you by first giving the exact element of $H_p$, say $H_p$ is abelian group of order $8$ given by $\{e,a_2,\cdots,a_8\}$, to make it a non-cyclic group (after some calculations or cayley table) $H_p=\{e, a_1, a_2, a_1a_2, b_1,a_1b_1, a_2b_1, a_1a_2b_1\}$ and $a_2^2=e$, $a_1^2=a_2$, $b_1^2=e$, and $a_i b_j = b_j a_i$ (he/she should agree with these, since this is an effort to make $H_p$ a non-cyclic abelian group of order 8, yes there might be other way). Since $a_1^2=a_2$, $H_p$ depends only on $a_1$ and $b_1$, so we can name $a=a_1$ and $b=b_1$ and hence
\[H_p = \{e,a,a^2,a^3,b,ab,a^2b\}\]
Do you realize? we've just stated
$$\{e,a_2,\cdots,a_8\}= \{e, a_1, a_2, a_1a_2, b_1,a_1b_1, a_2b_1, a_1a_2b_1\} = \{e,a,a^2,a^3,b,ab,a^2b\}$$
and we seem okay, though we throw $a_8$, $a_4$, $b_2$ etc. And we feel fine about that. We stated the three sets are the same, even though they are virtually not have the same elements. That's because we agree $a_1$ is $a$ and $b_1$ is $b$, and $a_2$ is $a_1^2$ and so on..
We agree! That is important in Mathematics.
Back on, what is actually "$=$" on proposition 1 and 2 ? In general it is isomorphism sign.
With some agreement, we can make it equal sign
(You know not everyone on this world use symbol $2$ to denote "two", in ancient roman its "II", in arabic it's different also, but are they really different in meaning? no, even sounded different in many languages, two is always two :D )
Hello formalism? :D :D
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