Let G be a group of order 100 that has a subgroup H of order 25. Prove that every element of G of order 5 is in H. I know that I have to use cosets, but I'm shaky on understanding cosets as it is. I was thinking of using the fact that G has a unique subgroup of order 25, but I don't see how that could work with cosets. 
Anyone have any ideas on how to get started?
 A: Let $H$ be the subgroup of $G$ of order $25$, and het $g$ be an element of order $5$ in $G$.
Consider the cosets: $$1H,gH,g^2H,g^3H,g^4H$$
These can't be disjoint, since that would mean $125$ distinct elements of $G$.
So $g^iH=g^jH$ for some $0\leq i <j<5$. Then $g^{j-i}H=H$, or $g^{k}\in H$ for some $1\leq k<5$. Now show that $g\in H$ by finding $m$ so that $g=(g^k)^m$.
A: If $x$ had order $5$ and were not in $H$, then the subgroup generated by $x$ and $H$ would have at least $125$ elements. But in $G$ there are only $100$.
A: If you know that $G$ has a unique subgroup of order $25,$ then note that $gHg^{-1}$ is a subgroup of $G$ of order $25$ for all $g\in G.$ Hence, $gHg^{-1}=H$ for all $g\in G,$ meaning that $H$ is a normal subgroup of $G.$ Now, suppose that there is some $g_0\in G$ of order $5$ such that $g_0\notin H.$ Note that we then have $g_0^n\in H$ if and only if $n$ is an integer multiple of $5.$ (Can you see why?) Thus, you can show that $H,g_0H,g_0^2H,g_0^3H,$ and $g_0^4H$ are distinct cosets of $H,$ so disjoint subsets of $G.$ But then the union of those cosets is also a subset of $G,$ and has cardinality $125,$ which is impossible, since $|G|=100.$
A: I like the simple direct approach of the others. More sophisticated approach would be applying Sylow theory: $[G:N_G(H)]$ must divide 1, 2 or 4 and $\equiv 1$ mod 5. Hence $G=N_G(H)$, that is $ H$ is normal and the unique Sylow $5$-Sylow subgroup. Then all elements of order $5$ must lie in $H$.
A: Let ${a \in G }$ and its order is 5, we know that set ${<a>H}$ has exactly ${ \frac {5 \cdot |H| }{|<a> \cap H|}}$ elements and ${|<a> \cap H|}$ divides ${|<a>| \implies}$ ${|<a> \cap H|=5}$ hence, ${<a> \cap H = <a> \implies a \in H}$
