A subgroup is a subset which is also a group of its own, in a way compatible with the original group structure. But not every subset is a subgroup. To be a subgroup you need to contain the …

Apr 10, 2013 · How does one prove that any proper subgroup of $A_5$ has order at most $12$? I have seen that there are $24$ $5$-cycles and $20$ $3$-cycles. What do the other members ...

Understanding how to prove when a subset is a subgroup Ask Question Asked 9 years, 4 months ago Modified 4 years, 2 months ago

Jan 31, 2025 · So, if $P$ and $Q$ are Sylow $5$ -subgroup and Sylow $7$ -subgroup of $G$ respectively, then one of the two has to be normal in $G$. Assume $P$ is normal in $G$, that …

Aug 10, 2012 · I've been reading some stuff about algebra in my free time, and I think I understand most of the stuff but I'm having trouble with the exercises. Specifically, the …

Suppose $Q_8$ is isomorphic to subgroup of $S_n$ with $n\leq 7.$ Then it should come from a group action of $Q_8$ on a set of cardinality at most 7. Suppose $Q_8$ acts on a set $A$ with …

A subgroup generated by a set is defined as (from Wikipedia): More generally, if S is a subset of a group G, then , the subgroup generated by S, is the smallest subgroup of G containing every e...

Jan 6, 2013 · Any finitely generated subgroup of $\mathbb Q$ must be cyclic and, and a subgroup of $\mathbb Q$ is isomorphic to atleast one subgroup of $\mathbb Q$ containing $1$. I think …

For all the subgroups on the third row from the top, their only proper subgroup is the trivial subgroup, which is trivially normal to $G$, so it doesn't make sense to use any of the …

Until recently, I believed that a subgroup of a direct product was the direct product of subgroups. Obviously, there exists a trivial counterexample to this statement. I have a question regarding...