The question is not much technical; I wanted to get to know the statement of a Theorem little clear. (I am not considering the proof of the Theorem).

Let $p$ always denote an odd prime, and $M_p$ is the non-abelian group of order $p^3$ and exponent $p$. If $|G|=p^n$, let $s_k(G)$ denote the number of subgroups of order $p^k$ in $G$ ($1\le k\le n$). Let $C_p^r$ denote the direct product of $r$ copies of cyclic group $C_p$ of order $p$.

Theorem 1.4: (see Ref. below) Let $|G|=p^n$, $p$ odd prime, and $\tilde{G}:=M_p\times C_p^{n-3}$. If G is not elementary abelian, then for $1\le k\le n$, we have $s_k(G)\le s_k(\tilde{G})$. In particular, if $2\le k\le n-2$, then $s_k(G)< s_k(\tilde{G})$.

Here, I didn't understand strict inequality in last part; I think, we must take $G\not\cong \tilde{G}$.

Q. The last part of the Theorem should be,

In particular, if $2\le k\le n-2$ ("and $G\not\cong \tilde{G}$"), then $s_k(G)<s_k(\tilde{G})$.

Is this correct? (More precisely, if $G\not\cong \tilde{G}$, is it true that we have strict inequality? I think, author want's to state this way; but I confused!)

I saw the papers where above Theorem is cited, and in the places of citation where the Theorem is stated completely, I still saw same statement, with no mention of $G\not\cong \tilde{G}$.

Ref: H. Qu, Finite non-elementary abelian p-groups whose number of subgroups is maximal, Israel Journal of Mathematics, June 2013, Volume 195, Issue 2, pp 773–781.

The question is not much technical; I wanted to get to know the statement of a Theorem little clear. (I am not considering the proof of the Theorem).

Let $p$ always denote an odd prime, and $M_p$ is the non-abelian group of order $p^3$ and exponent $p$. If $|G|=p^n$, let $s_k(G)$ denote the number of subgroups of order $p^k$ in $G$ ($1\le k\le n$). Let $C_p^r$ denote the direct product of $r$ copies of cyclic group $C_p$ of order $p$.

Theorem 1.4: (see Ref. below) Let $|G|=p^n$, $p$ odd prime, and $\tilde{G}:=M_p\times C_p^{n-3}$. If G is not elementary abelian, then for $1\le k\le n$, we have $s_k(G)\le s_k(\tilde{G})$. In particular, if $2\le k\le n-2$, then $s_k(G)< s_k(\tilde{G})$.

Here, I didn't understand strict inequality in last part; I think, we must take $G\not\cong \tilde{G}$.

Q. The last part of the Theorem should be,

In particular, if $2\le k\le n-2$ ("and $G\not\cong \tilde{G}$"), then $s_k(G)<s_k(\tilde{G})$. [Is this correct?]

Why it confused me? More precisely, if $G\not\cong \tilde{G}$, then the inequality is strict; I think, author want's to state this way. I saw the papers (after 2013) where above Theorem is cited; in one or two papers, it is the above result is mentioned, but not with the correction I am expecting in question. This kept me confused about precise statement of the Theorem.

Ref: H. Qu, Finite non-elementary abelian p-groups whose number of subgroups is maximal, Israel Journal of Mathematics, June 2013, Volume 195, Issue 2, pp 773–781.