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

[mathmari]

Hey!

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)

 

[Evgeny.Makarov]

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.

 

[I like Serena]

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)

 

[Deveno]

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 $|f(G_1)| \in \{1,3,5,15\} \cap \{1,2,4,7,8\} = \{1\}$.

 

[mathmari]

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 $b \neq e_2$.

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 $f(G_1) = \{e_2\}$??

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)

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??

 

[Deveno]

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?

 

[mathmari]

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)

So there is an isomorphism only when $|G_1|=|G_2|$ ?? (Wondering)

 

[Evgeny.Makarov]

So there is an isomorphism only when $|G_1|=|G_2|$ ?

Yes, this is a necessary condition.

 

[mathmari]

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)

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)

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 - - -

Yes, this is a necessary condition.

I see... Thanks a lot! (Smile)

 

[I like Serena]

So there is an isomorphism only when $|G_1|=|G_2|$ ?? (Wondering)

Yes, this is a necessary condition.

Yes. A necessary condition... and it is not a sufficient condition!

 

[mathmari]

Yes. A necessary condition... and it is not a sufficient condition!

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)

 

[I like Serena]

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.)

 

[Deveno]

Right! (Happy)

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

Fixed.

 

[I like Serena]

Fixed.

Heh. Doesn't $|G_1|=|G_2|$ imply they are finite?

 

[Deveno]

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.