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bham10246
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Question: Let [itex]G[/itex] be a group of order [itex]p^n > 1[/itex] where [itex]p[/itex] is prime. If [itex]H[/itex] is a subgroup of [itex]G[/itex], show that it is subnormal in [itex]G[/itex]. That is, I need to show that there is a chain of subgroups [itex]H=H_0 \triangleleft H_1 \triangleleft ...\triangleleft H_m = G[/itex], where [itex]m\leq n[/itex].
Analysis: We can easily show by induction on n that there is a series of normal subgroups of G
[itex]1=G_0 < G_1 < ... < G_n = G [/itex] such that [itex][G_{i+1}:G_i]=p[/itex].
Since H is a subgroup of G, [itex]|H|=p^k [/itex] where [itex]k \leq n[/itex]. Since H is a p-group, H also has a series of normal subgroups of H.
Perhaps if H is normal in G, then we can consider G/H which has a series of normal subgroups
[itex]\bar{1}=\bar{H_0}< \bar{H_1} < ... <\bar{H_m}=G/H[/itex].
By Fourth Isomorphism Theorem, does that mean we have a series of normal subgroups [itex]H=H_0 < H_1 < ...< H_m = G[/itex], where [itex]m\leq n[/itex]?
Firstly, how do we know that H is normal in G? :uhh:
Secondly, is [itex]H_i [/itex] normal in [itex]H_{i+1}[/itex]?
Thanks for your time!
Analysis: We can easily show by induction on n that there is a series of normal subgroups of G
[itex]1=G_0 < G_1 < ... < G_n = G [/itex] such that [itex][G_{i+1}:G_i]=p[/itex].
Since H is a subgroup of G, [itex]|H|=p^k [/itex] where [itex]k \leq n[/itex]. Since H is a p-group, H also has a series of normal subgroups of H.
Perhaps if H is normal in G, then we can consider G/H which has a series of normal subgroups
[itex]\bar{1}=\bar{H_0}< \bar{H_1} < ... <\bar{H_m}=G/H[/itex].
By Fourth Isomorphism Theorem, does that mean we have a series of normal subgroups [itex]H=H_0 < H_1 < ...< H_m = G[/itex], where [itex]m\leq n[/itex]?
Firstly, how do we know that H is normal in G? :uhh:
Secondly, is [itex]H_i [/itex] normal in [itex]H_{i+1}[/itex]?
Thanks for your time!
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