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Let $G$ be a group of order $5^k\cdot 8$.

I was trying to prove that there are normal subgroups of order $5^{k},5^{k}\cdot2,5^{k}\cdot4$.

I saw the following statement:

Let $P$ be a $p$-Sylow subgroup of $G$. Then obviously $|P|=5^k$.

Why the order of $P$ is $5^k$? I can't figure the connection between the $p$-Sylow subgroup and $5^k$. From the third Sylow theorem I know that $n_{2}|5^k$ and $n_2\equiv 2\,(mod\,2)$ but how does it help me?

Edit:

After understanding what I missed, I'll try to solve it:

From the first Sylow theorem we know that $G$ has a $5$-Sylow subgroup of order $5^k$. I'll prove that $P\triangleleft G$. On one hand $n_5=|G\,:\,N_G(P)|$ and on the other hand (third Sylow theorem) we get $n_5\equiv 1\,(mod\, 5)$. From Lagrange we know that $P$ is subgroup of $N_G(P)$ so we know that $5^k$ divides $N_G(P)$ so there is $s\in\mathbb{N}$ so $|N_G(P)|=5^k\cdot s$. We get

$$|G\,:\,N_G(P)|=\frac{5^k\cdot 8}{5^k\cdot s}=\frac{8}{s}$$

We conclude that $n_5$ divides $8$ and the remainder is $1$ while dividing by $5$ meaning $n_5=1$. So we have only one $5$-Sylow subgroup meaning $P\triangleleft G$.

What do you think?

Edit 2: I was wondering if we need it all? From the Sylow $III$ theorem we get $n_5|8$ and $n_5\equiv 1 \, (mod \,5)$. from $n_5|8$ we get $n_5\in \{1,2,4,8\}$ and from $n_5\equiv 1 \, (mod \,5)$ we get $n_5=1$ so there is only one $5$-Sylow group meaning $P\triangleleft G$. Is it correct?

Edit 3 (Last edit I promise!)

I will prove that $G$ has a normal subgroup of order $4\cdot 5^k$. By Lagrange theorem we get:

$$|G/P|=\frac{|G|}{|P|}=\frac{5^{k}\cdot 8}{5^{k}}=8$$

$8$ is a power of prime number so we $G/P$ has a subgroup of order $4$ and $2$ of the following format: $A/P$. From Lagrange:

$$|A|=|A/P|\cdot|P|=4\cdot5^{k}$$

And Also,

$$|G\,:\,A|=\frac{|G|}{|A|}=\frac{8\cdot5^{k}}{4\cdot5^{k}}=2$$

So $A$ is a subgroup of $G$ of order $4\cdot 5^k$ and index $2$, so its normal.

Now I wonder how to prove that $G$ has a normal group of order $2\cdot 5^k$. We stated that $G/P$ has a subgroup $B/P$ of order $2$. So from the Lagrange theorem we get:

$$|B|=|G/P|\cdot|P|=2\cdot5^{k}$$

I got stuck. I checked ther solution and found out that:

$G/P$ is of order $8$ so it has a subgroup of order $2$ in $Z(G/P)$. This group is of form $B/P$ for some subgroup $B$ of $G$. We know that $B/P\subset Z(G/P)$ so $B/P \triangleleft G/P$ meaning $B\triangle G$.

First of all, I don't understand why $G/P$ has a subgroup of order $2$ in $Z(G/P)$. I do know that $Z(G/P)\leq G/P$, but how does it help up. Also, I don't understand why $B/P\subset Z(G/P)$ and why $B/P \triangleleft G/P$?

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    $\begingroup$ The Sylow-5-subgroup has order $5^k$. $\endgroup$ – ancientmathematician Feb 18 at 16:54
  • $\begingroup$ You're misquoting Sylow III. $n_2$ is odd. $\endgroup$ – ancientmathematician Feb 18 at 16:55
  • $\begingroup$ And a final hint: can you do the case $k=0$? $\endgroup$ – ancientmathematician Feb 18 at 16:57
  • $\begingroup$ @ancientmathematician Uhh, I missed (for some reason) that $5$ is a prime (feel ashamed). I'll edit my question in order to show how I tried to solve it. Thank you. $\endgroup$ – vesii Feb 18 at 16:59
  • $\begingroup$ What can you say about $n_5$? $\endgroup$ – the_fox Feb 18 at 17:04
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Hint: Let $\overline G$ denote $G/P $. Since $\overline G$ is a $p$-group, $Z (\overline G) $ is non-trivial. The possible orders of $Z (\overline G) $ are $2,4$ and $8$. If the order is $2$, then we are done.

If the order is $4$ or $8$, then by Cauchy's theorem $Z (\overline G) $ has an element of order $2$. Now consider the subgroup generated by that element.

Also note that $Z(\overline G)$ is a normal subgroup of $\overline G$(why?).

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