Size of A4 =12

So, there are either 1 or 3 2-sylow subgroups

and there are either 1 or 4 3-sylow subgroups.

Suppose that there are 4 3-sylow subgroups. Then, there are 8 elements of order 3

Thus, 2-sylow subgroup must be unique.

However, this is case when I suppose that there are 3-sylow subgroups.

What If I suppose that there are 2-sylow subgroups and derive the result that there can't be 4 3-sylow subgroups?

Also, How can I find the subgroup of order 4?

Do I have to look at every element in A4 and consider possible cases?


You know that the Klein 4 group V=(1,(12)(34),(13)(24),(14)(23))is a subgroup of order 4 of A4. It's a normal subgroup of order 4 of A4. Any automorphism of A4 must map an element of order 2 to an element of order 2, and since V contains only the 3 elements of order 2 and the idendity, then V is stable under any automorphism of A4,, and hence is a normal subgroup of A4. V is a normal 2 sylow subgroup of A4, and hence is a unique 2 sylow subgroup (order 4) of A4.


Looking at the elements of $A_4$ is not that difficult. If you know what the elements of $S_4$ look like (identity, $2$-, $3$-, $4$-cycles and products of two disjoint transpositions (i.e. $2$-cycles)), pick the ones with even parity: $(1)$, $(a \ b \ c)$ and $(a \ b)(c\ d)$, which have orders $1$, $3$ and $2$ respectively. Since there are no elements of order $4$, each Sylow $2$-subgroup consists of the identity and $3$ elements of order $2$. It's not difficult to see that there are only $3$ elements of the form $(a \ b)(c\ d)$. Hence you can only form one subgroup of order $4$.


With inspection: H = {(1), (12)(34), (13)(24), (14)(23)} is a subgroup of order 4. Since a subgroup of order 4 would be permitted only elements of order 1, 2 or 4 and since the ones in H are the only ones with such order in G then H is the only subgroup of order 4.

You can't do what you are suggesting without consideration of what A4 means. Take for example D12 (dihedral group of order 12) which is not isomorphic to A4, this has 3 sylow 2 subgroups. I will leave it to the reader to verify this.


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