Group of order $5^k\cdot 8$ has normal subgroups of order $5^{k},5^{k}\cdot2,5^{k}\cdot4$

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 $$B/P \triangleleft G/P$$?

• The Sylow-5-subgroup has order $5^k$. – ancientmathematician Feb 18 at 16:54
• You're misquoting Sylow III. $n_2$ is odd. – ancientmathematician Feb 18 at 16:55
• And a final hint: can you do the case $k=0$? – ancientmathematician Feb 18 at 16:57
• @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. – vesii Feb 18 at 16:59
• What can you say about $n_5$? – the_fox Feb 18 at 17:04

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?).