Conjugation action Let G be a non-abelian group of order 10. Show that the conjugation action on the set of elements of order 2 induces an injective homomorphism from G to S5.
Can anyone give me some hints? 
 A: Your group has a (cyclic) subgroup of order 5, which is normal being of index 2. 
This subgroup does not have any element of order 2!
As conjugation action always preserves the complement of a normal subgroup, a set of 5 elements, you get a homomorphism to $S_5$. Check that they are all of order 2. (Actually this group is the group of symmetries of the regular pentagon; but you don't need to use it).
A: Hints:


*

*Since $G$ is non-abelian and of order $2 \cdot 5$, the subgroup of order $5$ (there always exists atleast one, why?) is a normal subgroup. (Why?)

*If you know Sylow's theorems, then the following hint is a easy shortcut:
What is the number of subgroups of order 2? (We know it is atleast 1, why? If it is one, then the group becomes abelian since it is the product of two normal subgroups. So this can't be the case. Then what does Sylow's theorem tell you?)


*If you can deduce there must be five subgroups (equivalently five elements) of order $2$, then the conjugation map permutes these among each other. (Since image of a subgroup under an automorphism has the same order) Thus, conjugation action on the set of elements of order 2 induces an injective homomorphism from $G$ to $S_5$.


Try to write the homomorphism explicitly. Also ask if more clarification is required.
A: Suppose $a$ is an element of order $2$ in $G$, and $b$ is an element of order $5$. If $A = \langle a\rangle$ and $B = \langle b\rangle$,
then $|AB| = \dfrac{|A||B|}{|A \cap B|} = \dfrac{2\cdot 5}{1} = 10 = |G|$.
This shows that $G = \langle a,b\rangle$. Explicitly, then:
$G = \{e,a,b,b^2,b^3,b^4,ab,ab^2,ab^3,ab^4\}$. Since $G$ is non-abelian, none of these have order $10$ (or else $G$ would be cyclic). It can be shown that $ba$ cannot be $e,a$ nor any power of $b$, so that $ba$ must be $ab^2,ab^3$ or $ab^4$ ($G$ non-abelian rules out $ab$). We can rule out the first two like so:
$ba = ab^2 \implies (ab)^2 = a(ba)b = a(ab^2)b = b^3$ so:
$(ab)^3 = (ab)(ab)^2 = (ab)b^3 = ab^4$
$(ab)^4 = ((ab)^2)^2 = b^6 = b$
$(ab)^5 = (ab)(ab)^4 = ab^2$, so that $ab$ has order $10$, contradiction.
Similar calculations with $ba = ab^3$ show that $ab$ would likewise have to be order $10$.
So the only viable option is $ba = ab^4 = ab^{-1}$, and this leads to:
$(ab)^2 = a(ba)b = a(ab^{-1})b = a^2 = e$. Since we no longer have room for another subgroup of order $5$, we have $5$ elements of order $2$, namely: $\{a,ab,ab^2,ab^3,ab^4\}$.
(Most of the above can be side-stepped in two lines if you know Sylow Theory).
Now $B$ is normal in $G$ (it's of index $2$), which means that if $x \not\in B$, neither is $gxg^{-1}$, for any $g \in G$ (for if $gxg^{-1} \in B$, then so is $x = g^{-1}(gxg^{-1})g$, by normality of $B$).
So we can let $G$ act on $\{a,ab,ab^2,ab^3,ab^4\}$ by conjugation:
$g \cdot ab^k = g(ab^k)g^{-1}$.
This induces a homomorphism $G \to S_5$ (simply relabel $a$ as $1$, $ab$ as $2$, etc.). For example, if we conjugate by $b$, we get:
$bab^{-1} = ab^3$
$b(ab)b^{-1} = ab^4$
$b(ab^2)b^{-1} = a$
$b(ab^3)b^{-1} = ab$
$b(ab^4)b^{-1} = ab^2$
So $b$ gets mapped under our homomorphism to $(1\ 4\ 2\ 5\ 3)$.
Since the image of $G$ contains an element of order $5$, the kernel of our homomorphism can have cardinality of at most $2$ (the homomorphism clearly isn't trivial, since it contains a non-trivial image element, and if the kernel had order $5$, the image would have order $2$ and could not have an element of order $5$).
We also have, conjugating by $a$:
$aaa^{-1} = a$
$a(ab)a^{-1} = ab^4$
$a(ab^2)a^{-1} = ab^3$
$a(ab^3)a^{-1} = ab^2$
$a(ab^4)a^{-1} = ab$, which corresponds to the permutation $(2\ 5)(3\ 4)$.
Hence the image of our homomorphism also contains an element of order $2$, which means our image has order divisible by $10$, and thus must be $10$. Hence the kernel of our homomorphism is trivial, and is our homomorphism is thus injective.
There are "slicker", and much more elegant ways to get to this place. I've written out the gory details to show that this is actually very "basic" one doesn't need the "high-powered" theorems, and hopefully the explicit demonstration of some actual image elements allows you to see the "nuts and bolts" of what is happening at the "element level".
My apologies in advance to any who feel this is "too much information".
