Why does it imply that if f is onto there are a ≠ e_1 with f(a) ≠ e_2 ?

  • MHB
  • Thread starter mathmari
  • Start date
In summary: Thinking)... subgroup of a group of order 56, because the only divisors of 15 are 1,3, and 5, and none of these numbers divide 56. Therefore, there can be no homomorphic image of $G_1$ that lives inside $G_2$. In summary, the conversation discusses a mathematical problem involving groups and isomorphisms. The problem states that for two groups $G_1$ and $G_2$ with sizes 15 and 56 respectively, it is not possible to have an isomorphism between them. The conversation then goes on to explain a proof of this statement, highlighting the properties of homomorphisms, subgroup orders, and Lagrange's
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

I am facing some difficulties understanding the solution of the exercise:

"Let $G_1,G_2$ groups. $|G_1|=15, |G_2|=56$
Show that it is not possible that there exist an isomorphism $f:G_1 \rightarrow G_2$."

which is the following:

($<$ means subgroup)
($e_1$:identity element of $G_1$, $e_2$:identity element of $G_2$)

$H_1=f^{-1}(G_2)<G_1$ from the property: If $H_2<G_2$, then $f^{-1}(H_2)<G_1$

$H_1$ is a subgroup of $G_1$ $\displaystyle{\overset{\text{Lagrange}}{\Rightarrow} |H_1| \mid 15}$

So $|H_1| \in \{1,3,5,15\}$

If $|H_1|=1$, that means that $H_1=\{e_1\}$, that means that $f^{-1}(G_2)=\{a \in G_1: f(a) \in G_2\}=\{e_1\}$.
But $f$ is onto, so there are $a \neq e_1$ with $f(a) \neq e_2$.
If it were $f^{-1}(G_2)=\{e_1\}$ that would mean that all $f(a)=e_1$.

So $|H_1| \in \{3,5,15\}$

Since $H_1<G_1$, $f(H_1)<G_2$, so $|f(H_1)| \mid |G_2|$

Since $f$ is $1-1$, $|f(H_1)|=3,5,15$

$|G_2|=56$ but none of the $3,5,15$ divide $56$.
Could you explain me the red part?? Why does this stand?? (Wondering)
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
mathmari said:
If $|H_1|=1$, that means that $H_1=\{e_1\}$, that means that $f^{-1}(G_2)=\{a \in G_1: f(a) \in G_2\}=\{e_1\}$.
But $f$ is onto, so there are $a \neq e_1$ with $f(a) \neq e_2$.
If it were $f^{-1}(G_2)=\{e_1\}$ that would mean that all $f(a)=e_1$.
Since $|G_2|>1$, there exists a $b\in G_2$ such that $b\ne e_2$, and since $f$ is onto, there must exist an $a\in G_1$ such that $f(a)=b\ne e_2$.

However, I don't understand the second red line. First, it probably should say that $f(a)=e_2$, not $f(a)=e_1$, since $f(a)\in G_2$. Second, $f^{−1}(G_2)=\{e_1\}$ does indeed imply $f(a)=e_1$ for all $a$, but not because there is any reasonable relationship between these two statements, but because $f^{−1}(G_2)=\{e_1\}$ is a contradiction. Indeed, $f$ maps all elements of $G_1$ into $G_2$, so the pre-image of $G_2$ must be all of $G_1$. Where do other elements of $G_1$ besides $e_1$ go if not $G_2$?

Finally, the whole problem is bizarre. An isomorphism is first of all a bijection, and there is no bijection between sets with sizes 15 and 56. All in all, I would say the proof is reta... well, weird.
 
  • #3
Evgeny.Makarov said:
Finally, the whole problem is bizarre. An isomorphism is first of all a bijection, and there is no bijection between sets with sizes 15 and 56. All in all, I would say the proof is reta... well, weird.

Regardless of the red stuff, I quite agree that an isomorphism between finite sets with different sizes is immediately a contradiction. (Wasntme)
 
  • #4
One of the things about isomorphic groups is that they share every "group property". It's a bit complicated to say which properties are "group properties" and which ones aren't, but here is one:

Isomorphic groups have the same order.

Since $|G_2| = 56 > |f(G_1)| = |G_1| = 15$, it's clear that $G_1$ and $G_2$ are NOT isomorphic.

We can, however, say even more:

If $f:G_1 \to G_2$ is any HOMOMORPHISM, then $f(G_1) = \{e_2\}$ (that is, $f$ is the TRIVIAL homomorphism which sends everything to the identity of $G_2$).

It is clear that $f(G_2)$ (the image of $f$) forms a subgroup of $G_2$, which by Lagrange, has order dividing 56. Since $f$ can be at MOST 1-1, the image of $f$ has at most 15 elements, and is a divisor of 56. So possible orders are: 1,2,4,7, and 8.

If we have $K = f^{-1}(e_2)$, this is a subgroup of $G_1$, and this subgroup has order a divisor of 15: so 1,3,5 or 15.

Finally, we have:

$|f(G_1)| = [G_1:K] = \dfrac{|G_1|}{|K|}$, so $|f(G_1)|\ast|K| = |G_1|$.

This shows that $|f(G_1)|$ divides the order of $G_1$, so $|f(G_1)| \in \{1,3,5,15\} \cap \{1,2,4,7,8\} = \{1\}$.
 
  • #5
Evgeny.Makarov said:
Since $|G_2|>1$, there exists a $b\in G_2$ such that $b\ne e_2$, and since $f$ is onto, there must exist an $a\in G_1$ such that $f(a)=b\ne e_2$.

Ahaa.. Ok!
Knowing that $f$ is onto, do we know that this $a$ is $\neq e_1$ ?? (Thinking)

Is it as followed??

Since there is more than one element in $G_2$, for example $\{b,e_2\}$, there must be at least two elements in $G_1$ for example $\{a, e_1\}$, right??
And since $f$ is an homomorphism, $f(e_1)=e_2$.
So, $f(a)=b$, where $a \neq e_1$ and $b \neq e_2$.

Deveno said:
If $f:G_1 \to G_2$ is any HOMOMORPHISM, then $f(G_1) = \{e_2\}$ (that is, $f$ is the TRIVIAL homomorphism which sends everything to the identity of $G_2$).

I got stuck right now... (Worried)

Why does it stand that if $f:G_1 \to G_2$ is an homomorphism, then $f(G_1) = \{e_2\}$??

Deveno said:
It is clear that $f(G_2)$ (the image of $f$) forms a subgroup of $G_2$, which by Lagrange, has order dividing 56. Since $f$ can be at MOST 1-1, the image of $f$ has at most 15 elements, and is a divisor of 56. So possible orders are: 1,2,4,7, and 8.

Why does the image of $f$ have at most $15$ elements?? (Wondering)

$f:G_1 \to G_2$, $f$ is $1-1$ and we know that $|G_1|=15$

Does this imply that $f(G_2)$ has at most $15$ elements?? (Thinking)

Deveno said:
If we have $K = f^{-1}(e_2)$, this is a subgroup of $G_1$, and this subgroup has order a divisor of 15: so 1,3,5 or 15.

Finally, we have:

$|f(G_1)| = [G_1:K] = \dfrac{|G_1|}{|K|}$, so $|f(G_1)|\ast|K| = |G_1|$.

This shows that $|f(G_1)|$ divides the order of $G_1$, so $|f(G_1)| \in \{1,3,5,15\} \cap \{1,2,4,7,8\} = \{1\}$.

So the only possible subgroup is the one with order $1$, which is $f(G_1) = \{e_2\}$, right??
And why can this not stand??
 
  • #6
mathmari said:
I got stuck right now... (Worried)

Why does it stand that if $f:G_1 \to G_2$ is an homomorphism, then $f(G_1) = \{e_2\}$??

Not for ANY two groups, for these two groups in this problem.
Why does the image of $f$ have at most $15$ elements?? (Wondering)

$f:G_1 \to G_2$, $f$ is $1-1$ and we know that $|G_1|=15$

Does this imply that $f(G_2)$ has at most $15$ elements?? (Thinking)

Yes, we can only have as many images as domain elements (although if $f$ is NOT injective, we will have fewer).

So the only possible subgroup is the one with order $1$, which is $f(G_1) = \{e_2\}$, right??
And why can this not stand??

It CAN "stand", our conclusion is that this is the ONLY possible homomorphism between a finite group of order 15, and a finite group of order 56. Basically, a homomorphic image of $G_1$ has to live inside $G_2$, and this homomorphic image has to have order dividing 15 and 56. But gcd(15,56) = 1.

Another way to say this is: no group of order 15 has a quotient group that is a subgroup of order 56, except for the trivial quotient: $G_1/G_1$.

It turns out that information about INTEGERS, tells us something about groups: a lot of the structure of groups has to do with the factorization of the group's order. You will see this later when you learn about Sylow theorems.

Already you should know that a group of order $p$, a prime, is cyclic. Can you tell me why?
 
  • #7
Evgeny.Makarov said:
An isomorphism is first of all a bijection, and there is no bijection between sets with sizes 15 and 56. All in all, I would say the proof is reta... well, weird.

I like Serena said:
Regardless of the red stuff, I quite agree that an isomorphism between finite sets with different sizes is immediately a contradiction. (Wasntme)

So there is an isomorphism only when $|G_1|=|G_2|$ ?? (Wondering)
 
  • #8
mathmari said:
So there is an isomorphism only when $|G_1|=|G_2|$ ?
Yes, this is a necessary condition.
 
  • #9
Deveno said:
Not for ANY two groups, for these two groups in this problem.

Yes, we can only have as many images as domain elements (although if $f$ is NOT injective, we will have fewer).

Ahaa..Ok! (Smile)
Deveno said:
It CAN "stand", our conclusion is that this is the ONLY possible homomorphism between a finite group of order 15, and a finite group of order 56. Basically, a homomorphic image of $G_1$ has to live inside $G_2$, and this homomorphic image has to have order dividing 15 and 56. But gcd(15,56) = 1.

I got stuck right now... Does the fact that it is the only possible homomorphism between a finite group of order 15, and a finite group of order 56, imply that it is not possible that there exist an isomorphism?? (Wondering)
Deveno said:
Already you should know that a group of order $p$, a prime, is cyclic. Can you tell me why?

The reason is the following, isn't it??

$G$ is a group of order $p$, where $p$ is a prime.
Let $a$ an element of $G$, $a \neq e$.
Then the cyclic subgroup of $G$ that is generated by $a$ has at least two elements, $a$ and $e$.
According to the Lagrange's Theorem, the order $m \geq 2$ of $<a>$ must divide $p$.
So it must be $m=p$.
So $<a>=G$, so $G$ is cyclic.

- - - Updated - - -

Evgeny.Makarov said:
Yes, this is a necessary condition.

I see... Thanks a lot! (Smile)
 
  • #10
mathmari said:
So there is an isomorphism only when $|G_1|=|G_2|$ ?? (Wondering)

Evgeny.Makarov said:
Yes, this is a necessary condition.

Yes. A necessary condition... and it is not a sufficient condition! :rolleyes:
 
  • #11
I like Serena said:
Yes. A necessary condition... and it is not a sufficient condition! :rolleyes:

So, when $|G_1| \neq |G_2|$ then we know that there is no isomorphism.
When $|G_1|=|G_2|$, then we have to define a map $f: G_1 \rightarrow G_2$ and show that $f$ is an homomorphism, $1-1$ and onto, right?? (Thinking)
 
  • #12
mathmari said:
So, when $|G_1| \neq |G_2|$ then we know that there is no isomorphism.
When $|G_1|=|G_2|$, then we have to define a map $f: G_1 \rightarrow G_2$ and show that $f$ is an homomorphism, $1-1$ and onto, right?? (Thinking)

Right! (Happy)

(And to be fair, either of 1-1 or onto suffices.)
 
  • #13
I like Serena said:
Right! (Happy)

(And to be fair, either of 1-1 or onto suffices...when $|G_1|$ is finite)

Fixed.
 
  • #14
Deveno said:
Fixed.

Heh. Doesn't $|G_1|=|G_2|$ imply they are finite?
 
  • #15
No. For example: $|\Bbb Z| = |\Bbb Q|$, but these groups are not isomorphic.

Orders need not be finite, and $|G|$, for a group $G$, is the cardinality of the underlying set, which also need not be finite.

We can conclude, for example, that no finite group is isomorphic to ANY infinite group.
 

FAQ: Why does it imply that if f is onto there are a ≠ e_1 with f(a) ≠ e_2 ?

Why does it imply that if f is onto there are a ≠ e1 with f(a) ≠ e2 ?

This implication is based on the definition of onto functions. An onto function, also known as a surjective function, is a function where every element in the codomain has at least one preimage in the domain. In other words, for every element e2 in the codomain, there exists at least one element a in the domain such that f(a) = e2. Therefore, if there exists an e1 in the codomain that does not have a preimage in the domain, this contradicts the definition of an onto function.

Can you provide an example of an onto function where a ≠ e1 with f(a) ≠ e2 ?

One example of an onto function that satisfies this implication is the function f: ℝ → ℝ defined by f(x) = x2. In this case, e1 = -1 does not have a preimage in the domain (-∞, ∞) as there is no real number that, when squared, equals -1. This function satisfies the definition of an onto function since every real number has at least one preimage (e.g. f(2) = 4, f(-2) = 4, etc.).

What if there are multiple elements in the domain that map to the same element in the codomain?

In this case, it is still possible for the implication to hold. For example, let f: ℤ → ℤ be defined by f(x) = x2. In this function, there are multiple elements in the domain that map to the same element in the codomain (e.g. f(2) = f(-2) = 4). However, there still exists an element in the codomain (e.g. e1 = -1) that does not have a preimage in the domain.

Is it possible for the implication to hold if f is not onto?

No, if f is not onto, then there exists at least one element in the codomain that does not have a preimage in the domain. Therefore, the implication cannot hold as there is no element a ≠ e1 in the domain that maps to e2 in the codomain.

How does this implication relate to the injectivity of a function?

This implication does not directly relate to the injectivity of a function. A function can be onto without being injective, and vice versa. However, if a function is both onto and injective, then it is known as a bijective function and the implication holds. In other words, if a function is bijective, then for every element e2 in the codomain, there exists exactly one element a in the domain such that f(a) = e2.

Similar threads

Back
Top