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Barre
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Homework Statement
If G is a finite abelian group and p is a prime such that [itex]p^n [/itex] divides order of G, then prove that G has a subgroup of order [itex]p^n[/itex]
Homework Equations
Theorem of Finite Abelian Groups: Every finite abelian group G is a direct sum of cyclic groups, each of prime power order.
The Attempt at a Solution
I of course have worked on a solution myself, and it's simply the last step that I'm unsure about.
Since G is finite, by Fundamental Theorem of Abelian Groups we know that
[itex]G = C_1 \times C_2 \times ... \times C_t[/itex]
for a finite number of cyclic groups [itex]C_i[/itex] of prime power order. Also, order of G and it's direct summands share the following relation:
[itex]|G| = |C_1||C_2|...|C_t| [/itex], and since [itex]p^n [/itex] divides |G|, there must be a finite set of summands such that product of their orders is equal to exactly [itex]p^n [/itex]. Let [itex]C_{p_1}, C_{p_2}, ... ,C_{p_m} [/itex] be those summands. The set:
[itex] S = \{(e_1, e_2,...,a_{p_1},...,a_{p_2},...,a_{p_m},..., e_t) | a_{p_i} \in C_{p_i} \} [/itex],
where all [itex]e_i [/itex] are the identity elements, is clearly a subgroup of [itex]C_1 \times C_2 \times ... \times C_n [/itex], of order [itex]|C_{p_1}||C_{p_2}| ... |C_{p_m}| = p^n [/itex].
Is it now OK to assume that since [itex]G = C_1 \times C_2 \times ... \times C_t [/itex], that G has a subgroup of order [itex]p^n [/itex] isomorphic to the subgroup S ?
It seems logical to me that this is the case, but I need confirmation. Also, is there a more obvious way of proving it? Of course all of this is just a consequence of Sylow theorems, but these come in the next section, so all I want to use is very elementary group theory + the theorem I described.
I'm also sorry about the set S, it might be totally unreadable but I had no idea how to otherwise describe it :D