Here is the question I am trying to answer:
Show that $A_{5}$ is a simple group. Do so in the following steps:
$(a)$ Write out all partitions of $5.$ Which of these correspond exclusively to even permutations?
$(b)$ Show that $A_{5}$ is generated by $3-$cycles.
Hint: It suffices to show that every product $(a b)(c d)$ of two transpositions is a product of $3-$cycles.
$(c)$ Show that, if $H \subset A_{5}$ is a normal subgroup and $H \neq \{e\},$ then $H$ contains a $3-$cycle.
$(d)$ Show that $H$ of part $(c)$ contains all $3-$cycles.
My question is:
I know the proofs of $a,b,c$ but the proof of $d$ was not clear for me, so I found this proof online:
" Now, We want to show that $H$ of part $(c)$ contains all $3-$cycles.\
By all cases of letter$(c)$ above, we have that $H$ contains a $3-$cycle, say $(a,b,c).$ Let $(u v w)$ be another $3-$cycle. Define $ g = \begin{pmatrix} a & b & c \\ u & v & w \\ \end{pmatrix}$ then $g(abc)g^{-1} = (u v w).$ Now, distinguish between 2 cases:
1 - If $g \in A_{5}$ and given $H$ is normal subgroup of $A_{5},$ then $g(abc)g^{-1} = (u v w) \in H.$
2- If $g \notin A_{5}$ then $g(ed) \in A_{5}$ as $A_{5}$ contains 5 letters and so $a,b,c,d,e$ are distinct. And so $g(ed)(abc)(g(ed))^{-1} \in H.$ Which can be written as $g(abc)(ed)(ed)g^{-1} \in H.$ And hence $g(abc)g^{-1}= (u v w) \in H$ because $(ed)(ed)$ is the identity permutation."
But I do not understand in the second case the following:
1- Why if $g \notin A_{5}$ then $g(ed) \in A_{5}$ ? the solution said "as $A_{5}$ contains 5 letters and so $a,b,c,d,e$ are distinct." but I do not see why this leads to that $g(ed) \in A_{5},$ could anyone explain this for me please?
2- Also, I do not understand in the second line in case 2, why $g(ed)(abc) = $g(abc)(ed),$ could anyone explain this for me please?
1 means: If $g$ is odd then $g(ed)$ is even which is obvious by definition of odd and even.
2 is true because $(ed), (a,b,c)$ commute being disjoint cycles.